Permutations and Combinations

To find the exponent of a prime $p$ in $n!$, we use:

$$\text{Exponent of } p \text{ in } n! = \sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor$$

Here, $p = 3$, and $n = 50$. So,

$$\sum_{k=1}^{\infty} \left\lfloor \frac{50}{3^k} \right\rfloor = \left\lfloor \frac{50}{3} \right\rfloor + \left\lfloor \frac{50}{9} \right\rfloor + \left\lfloor \frac{50}{27} \right\rfloor + \left\lfloor \frac{50}{81} \right\rfloor + \dots$$

Now compute:

• $\left\lfloor \frac{50}{3} \right\rfloor = 16$
• $\left\lfloor \frac{50}{9} \right\rfloor = 5$
• $\left\lfloor \frac{50}{27} \right\rfloor = 1$
• $\left\lfloor \frac{50}{81} \right\rfloor = 0$

Adding up:

$$16 + 5 + 1 = 22$$

Group -A Group -B

4B + 1G OB + 3G
3B + 2G 1B + 2G
2B + 3G 2B + 1G

$$\begin{gathered} Total possible cases \\ ={}^7\mathrm{C}_4\times {}^3\mathrm{C}_1\times {}^6\mathrm{C}_0\times {}^5\mathrm{C}_3+{}^7\mathrm{C}_3\times {}^3\mathrm{C}_2\times {}^6\mathrm{C}_1 \\ \times {}^5\mathrm{C}_2+{}^7\mathrm{C}_2\times {}^3\mathrm{C}_3\times {}^6\mathrm{C}_2\times {}^5\mathrm{C}_1 \\ =\frac{7\times 6\times 5}{3\times 2}\times 3\times 1\times \frac{5\times 4}{2}+\frac{7\times 6\times 5}{3\times 2} \\ \times 3\times 6\times \frac{5\times 4}{2}+\frac{7\times 6}{2}\times 1\times \frac{6\times 5}{2}\times 5 \\ = 1050 + 6300 + 1575 \\ = 8925 \end{gathered}$$

\( P_n=\) number of triangles formed by joining the vertices of an \(n -\)side regular polygon.

\( P_n={ }^n C_3 \)
\( { }^{n+1} C_3-{ }^n C_3=66\)
\( \Rightarrow \frac{(\mathrm{n}+1)(\mathrm{n})(\mathrm{n}-1)}{6}-\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)}{6}=66\)
\( \Rightarrow \mathrm{n}(\mathrm{n}-1)[(\mathrm{n}+1)-(\mathrm{n}-2)]=396 \)
\( \Rightarrow 3 \mathrm{n}(\mathrm{n}-1)=396\)
\( \mathrm{n}^2-\mathrm{n}-132=0 \)
\( \Rightarrow(\mathrm{n}-12)(\mathrm{n}+11)=0\)
\( \Rightarrow \mathrm{n}=12,-11 \)
\( 12=2^2 3^1\)
divisor are \(=1,2,3,4,6,12\)
prime divisor are \(=3,2 \)
sum \(=5\)

${40}^n=2^{3n}\times 5^n$

Now, highest power of 2 which divides \(60!\) is

$E_2\left(60!\right)=\left[\frac{60}{2}\right]+\left[\frac{60}{2^2}\right]+\left[\frac{60}{2^3}\right]+\left[\frac{60}{2^4}\right]+\left[\frac{60}{2^5}\right]$

$=30+15+7+3+1=56$

Now, highest power of 5 which divides \(60!\) is

$E_5\left(60!\right)=\left[\frac{60}{5}\right]+\left[\frac{60}{5^2}\right]$

\(=12+2=14\)

${40}^n={\left(2^3\right)}^n\times 5^n={\left(2^3\times 5\right)}^n$

$60!=2^{56}\times 5^{14}\dots =2^{14}\cdot {\left(2^3\cdot 5\right)}^{14}$

Maximum value of \(n\) is \(14\).

Dictionary order is \(A,D, P, R, U, Y\)

number of words starting with $A\to \frac{6!}{2!}=360$

number of words starting with $D\to \frac{6!}{2!}=360$

number of words starting with $P\to \frac{6!}{2!}=360$

number of words starting with $R\to \frac{6!}{2!}=360$

number of words starting with \(U A \rightarrow 5!=120\)

number of words starting with $UDAP\to 3!=6$

number of words starting with $UDAR\to 3!=6$

number of words starting with $UDAU\to 3!=6$

number of words starting with $UDAYPRU\to 1$

number of words starting with $UDAYPUR\to 1$

Total \(=1580\)

Required number of ways=

\begin{aligned}
& 8!-6!\times 3! \\
& =36,000
\end{aligned}

$$\begin{gathered} \text{Required number of ways}= \\ \text{Total ways} - \text{ways in which} {\mathrm{B}}_1 \text{and} {\mathrm{B}}_2 \text{are together} \\ =2!(3!4!)-2!(3!(3!2!\left)\right)=144 \end{gathered}$$

Total ways of selecting \(3\) points out of \(18\) is \({ }^{18} \mathrm{C}_3\).
Total ways of selecting \(3\) points from \(\mathrm{P}_1, \mathrm{P}_2, \mathrm{P}_3, \mathrm{P}_4, \mathrm{P}_5\) is \({ }^5 \mathrm{C}_3\).
Total ways of selecting \(3\) points from \(\mathrm{P}_6, \mathrm{P}_7, \ldots, \mathrm{P}_{11}\) is \({ }^6 \mathrm{C}_3\).
Total ways of selecting \(3\) points from \(\mathrm{P}_{12}, \mathrm{P}_{13}, \ldots, \mathrm{P}_{18}\) is \({ }^7 \mathrm{C}_3\).
Hence,The number of triangles, that can be formed using the points \(\mathrm{P}_1, \mathrm{P}_2, \ldots, \mathrm{P}_{18}\) is

\({ }^{18} C_3-{ }^5 C_3-{ }^6 C_3-{ }^7 C_3 \)
\(=751\)

The possible selections of balls are:
\((2\) blue, \(2\) yellow, \(4\) red), ( \(2\) blue, \(4\) yellow, \(2\) red),

( \(4\) blue, \(2\) yellow, \(2\) red), ( \(3\) blue, \(3\) yellow, \(2\) red),

( \(3\) blue, \(2\) yellow, \(3\) red) or ( \(2\) blue, \(3\) yellow, \(3\) red)
\(={ }^5 C_2 \cdot{ }^6 C_2 \cdot{ }^4 C_4+{ }^5 C_2 \cdot{ }^6 C_4 \cdot{ }^4 C_2+{ }^5 C_4 \cdot{ }^6 C_2 \cdot{ }^4 C_2+{ }^5 C_3 \cdot{ }^6 C_3\). \({ }^4 C_2+{ }^5 C_3 \cdot{ }^6 C_2 \cdot{ }^4 C_3+{ }^5 C_2 \cdot{ }^6 C_3 \cdot{ }^4 C_3\)

\(=150+900+450+1200+600+800 =4100\)

If number is divisible by 4 & 9 then is divisible by 36.

Total 3 digit numbers are 900

Required 3 digit number = Total 3 digit number which are divisible by 6 - total 3 digit number which are divisible by 36

= \(\frac{900}{6} - \frac{900}{36}\) = 125

Dictionary order is \(A,D, P, R, U, Y\)

number of words starting with $A\to \frac{6!}{2!}=360$

number of words starting with $D\to \frac{6!}{2!}=360$

number of words starting with $P\to \frac{6!}{2!}=360$

number of words starting with $R\to \frac{6!}{2!}=360$

number of words starting with \(U A \rightarrow 5!=120\)

number of words starting with $UDAP\to 3!=6$

number of words starting with $UDAR\to 3!=6$

number of words starting with $UDAU\to 3!=6$

number of words starting with $UDAYPRU\to 1$

number of words starting with $UDAYPUR\to 1$

Total \(=1580\)

Total number of \(\Delta\) are

\(={ }^9 \mathrm{C}_1{ }^{12} \mathrm{C}_2+{ }^9 \mathrm{C}_2{ }^{12} \mathrm{C}_1+{ }^1 \mathrm{C}_1{ }^9 \mathrm{C}_1{ }^{12} \mathrm{C}_1 \)
\(=594+432+108 \)
\(=1134\)

$\because$ No. of triangles having no side common with an sided polygon

$=\frac{{}^n{C}_1{.}^{n-4}{C}_2}{3}$

$=\frac{{}^8{C}_1{.}^4{C}_2}{3}=16$

$\because$ No. of triangles having no side common with an sided polygon

$=\frac{{}^n{C}_1{.}^{n-4}{C}_2}{3}$

$=\frac{{}^8{C}_1{.}^4{C}_2}{3}=16$

MATHS has only 5 letters, so in a 6-letter word at least one letter has to repeat.

Case 1: Exactly one letter is repeated twice
= ${}^{5}C_1·\frac{6!}{2!}=5\times 360=1800$

Case 2: Exactly two letters are repeated twice
=${}^{5}C_2·{}^{3}C_2\frac{6!}{2!2!}=5400$

Case 3: Exactly 3 letters are repeated twice =${}^{5}C_3\frac{6!}{2!2!2!}=900$

Hence the required solution = $1800+5400+900=8100$

$S=\left\{1,2,3,\dots .,9\right\}$

case-I: only one digit is repeated

$x={ }^9{C}_1\cdot { }^8{C}_7\times \frac{9!}{2}=\frac{9\times 8\times 9!}{2}$

case-II: two digits repeated

$y={ }^9{C}_2\cdot { }^7{C}_5\times \frac{9!}{2!\times 2!}=\frac{9\times 8}{2}\times \frac{7\times 6}{2}\times \frac{9!}{2!\times 2!}$

$\Rightarrow \frac{x}{y}=\frac{4}{21}$

$\Rightarrow 21x=4y$

\(\begin{aligned}
&\text { Correct option is (1) } \mathbf{1 6 8 0}\\
&\begin{aligned}
& \left\{2,2^2, 2^3 \ldots \ldots .^9\right\} \\
& \Rightarrow \frac{9!}{3!3!3!3!} \times 3!=\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4}{6 \times 6} \\
& =1680
\end{aligned}
\end{aligned}\)

$$\binom{22}{3} = \frac{22 \times 21 \times 20}{6} = 1540$$

Subtract the number of collinear sets of 3 points.

There are 13 points on line $L_1$: $O, P_1, ..., P_{12}$

From these 13 collinear points, number of 3-point combinations (collinear):

$$\binom{13}{3} = \frac{13 \times 12 \times 11}{6} = 286$$

There are 10 points: $O, Q_1, ..., Q_9$

From these 10 collinear points, number of 3-point combinations (collinear):

$$\binom{10}{3} = \frac{10 \times 9 \times 8}{6} = 120$$

Total number of non-degenerate triangles:

$\left(\genfrac{}{}{0px}{}{22}{3}\right)-\left(\genfrac{}{}{0px}{}{13}{3}\right)-\left(\genfrac{}{}{0px}{}{10}{3}\right)$$=1540-286-120=\boxed{1134}$

Given word, "KANPUR"

Words starting with \(A = 5!=120\)

Words starting with \(K = 5!=120\)

Words starting with \(N = 5!=120\)

Words starting with \(PA = 4!=24\)

Words starting with \(PK = 4!=24\)

Words starting with \(PN = 4!=24\)

Words starting with \(PRA = 3!=6\)

Next word is \(PRKANU = 1\)

FInal word is \(PRKAUN = 1\)

Total sum \(= 440\)

Hence, \(PRKAUN\) is the ${440}^{th}$ word

Total points are 12 on the three sides of triangle of which 3 points are collinear on side PQ,

4 points are collinear on side QR, 5 points are collinear on side PR,

Hence total number of triangles formed

$$\begin{gathered} ={}^{12}C_3-{}^{3}C_3-{}^{4}C_3-{}^{5}C_3 \\ =220-1-4-10 \\ =205 \end{gathered}$$

\(\left\{2,2^2,2^3,\ldots,2^9\right\}\)

\(\Rightarrow \frac{9!}{3!3!3!3!}\times 3!=\frac{9\times 8\times 7\times 6\times 5\times 4}{6\times 6}\)

\(=1680\)

$$\begin{gathered} \mathrm{Case} 1: 1,1,1,1,1,3,3 \to \mathrm{Total} 7 \mathrm{possible} \mathrm{numbers} \\ \mathrm{then} \mathrm{total} \mathrm{number} \mathrm{of} \mathrm{arrangments} \mathrm{are} \frac{7!}{5!.2!}=21 \\ \mathrm{Case} 2: 1,1,1,1,2,2,3 \to \mathrm{Total} 7 \mathrm{possible} \mathrm{numbers} \\ \mathrm{total} \mathrm{arrangement} \mathrm{will} \mathrm{be}= \frac{7!}{4!.2!} =105 \\ \mathrm{Case} 3: 1,1,1,2,2,2,2 \to \mathrm{Total} 7 \mathrm{possible} \mathrm{numbers} \\ \mathrm{total} \mathrm{arrangement} \mathrm{for} \mathrm{this} \mathrm{case} =\frac{7!}{4!.3!}=35 \\ \mathrm{So} \mathrm{total} \mathrm{number} \mathrm{of} \mathrm{required} \mathrm{arrangments} = 21+105+35 = 161 \end{gathered}$$

Total number of points are 12 in which 5 are collinear, Hence

No of triangles\(={ }^{12} \mathrm{C}_3-{ }^5 \mathrm{C}_3\)
\(= 220–10\)
\(= 210\)

Let \(a+b=22-2c\)

Since \(a\) and \(b\) must be non-negative integers ( \(a, b \geq 0\) ), their sum must also be non-negative:

\(22-2 c \geq 0 \Longrightarrow 2 c \leq 22 \Longrightarrow c \leq 11\)

Since \(c\) is a non-negative integer, the possible values for \(c\) are:

\(c \in\{0,1,2, \ldots, 11\}\)

For any fixed value of \(c\), let \(22-2 c=k\). The equation becomes:

\(a+b=k\)

The number of non-negative integral solutions ( \(a, b\) ) for the equation \(a+b=k\) is given by \({}^{k+2-1}C_{2-1}=k+1\)

Thus, for each value of \(c\), there are \((22-2 c)+1=23-2 c\) solutions.

Number of solutions of the equation

Total Solutions \(=\sum_{c=0}^{11}(23-2 c)=23 \times 12-2 \times \frac{11 \times 12}{2} \)
\(=12(23-11)=12^2 \)
\(=144\)

Total number of ways of arranging \(4\) boys and \(3\) girls in a queue \(=7!\)
Number of ways in which all \(3\) girls are together \(=3!\times 5!\)

because the \(3\) girls form one block and this block with \(4\) boys gives \(5\) objects

Required number of ways \(=7!-3!\times 5!\)

\(=5040-6 \times 120=5040-720=4320\)

Exponent of \(p\) in$n!=\sum _{k=1}^{\mathrm{\infty }}\lfloor \frac{n}{p^k}\rfloor$

Here, $p = 3$, and $n = 50$. So,

$\sum _{k=1}^{\mathrm{\infty }}\lfloor \frac{50}{3^k}\rfloor =\lfloor \frac{50}{3}\rfloor +\lfloor \frac{50}{9}\rfloor +\lfloor \frac{50}{27}\rfloor +\lfloor \frac{50}{81}\rfloor +\dots$

Now

$\left\lfloor \frac{50}{3} \right\rfloor = 16$

$\left\lfloor \frac{50}{9} \right\rfloor = 5$

$\left\lfloor \frac{50}{27} \right\rfloor = 1$

$\left\lfloor \frac{50}{81} \right\rfloor = 0$

Thus, $16+5+1=22$

If number is divisible by 4 & 9 then is divisible by 36.

Total 3 digit numbers are 900

Required 3 digit number = Total 3 digit number which are divisible by 6 - total 3 digit number which are divisible by 36

= \(\frac{900}{6} - \frac{900}{36}\) = 125

First, we need to arrange 5 girls. Which can be done in 5! ways.

o G o G o G o G o G o

There are 6 places which are to be filled by 3 boys.

This can be done in 6C, ways. Also, 3 boys can arrange themselves in 3! ways.

Required number ways

= 5! × ⁶C₃ × 3! = 14,400

C-I \((3G\) and \(2B)\) and \((1G\) and \(2B)\)

C-II \((2G\) and \(3B)\) and \((2G\) and \(1B)\)

C-III \((1G\) and \(4B)\) and \((3G\) and \(0B)\)

Total \(=\) C-I \(+\) C-II \(+\) C-III

\(={}^{7}C_2\cdot{}^{3}C_3\cdot{}^{6}C_2\cdot{}^{5}C_1+{}^{7}C_3\cdot{}^{3}C_2\cdot{}^{6}C_1{}^{5}C_2+{}^{7}C_4\cdot{}^{3}C_1\cdot{}^{6}C_0\cdot{}^{5}C_3\)

\(=8925\)

\(\sum_{r=1}^{30} \frac{r^2{ }^{30} c_r \cdot{ }^{30} C_r}{{ }^{30} C_{r-1}}=\sum_{r=1}^{30} r^2\left(\frac{30-r+1}{r}\right)\cdot{ }^{30} C_r\)

\(=\sum_{r=1}^{30} r(31-r) \cdot{ }^{30} C_r\)

\(=31 \sum_{r=1}^{30} r^{30} c_r-\sum_{r=1}^{30} r^2 \cdot c_r\)

\(=31\left(30 \cdot 2^{29}\right)-30(31) \cdot 2^{28}\)

\(=31 \times 15 \cdot 2^{29}\)

Total \(-\) when \(B_1\) and \(B_2\) are together

\(=2!(3!4!)-2!(3!(3!2!))=144\)

AA, MM, TT, H, I, C, S, E
Case (1): All distinct

\({ }^8 \mathrm{C}_5 \rightarrow 56\)

Case (2): 2 same, 3 different

\({ }^3 \mathrm{C}_1 \times{ }^7 \mathrm{C}_3 \rightarrow 105\)

Case (3): \(2\) same \(I^{\text {st }}\) kind, \(2\) same \(2^{\text {nd }}\) kind, \(1\) different

\({ }^3 \mathrm{C}_2 \times{ }^6 \mathrm{C}_1 \rightarrow 18\)

Total \(\rightarrow 179\)

Word starts from letter A will be 5! =120

Word starts from letter K will be 5! = 120

Word starts from letter N will be 5! = 120

Word starts from letter PA will be 4! = 24

Word starts from letter PK will be 4! = 24

Word starts from letter PN will be 4! = 24

Word starts from letter PRA will be 3! = 6

Total number of words = 438

then 439th word will be PRKANU

and 440th word will be PRKAUN

number of numbers formed using digits $\left(1111133\right)=\frac{7!}{5!2!}=21$

number of numbers formed using digits $\left(1111223\right)=\frac{7!}{4!2!}=105$

number of numbers formed using digits $\left(1112222\right)=\frac{7!}{4!3!}=35$

Total numbers $=161$

Here $\left(\genfrac{}{}{0px}{}{n}{r}\right)={}^n{C}_r$

Number of triangles: $p=\left(\genfrac{}{}{0px}{}{n}{3}\right)$
Number of quadrilaterals: $q = \binom{n}{4}$

Using the identity $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}$:

$p + q = \binom{n}{3} + \binom{n}{4} = \binom{n+1}{4} = 126$

$\frac{(n+1)n(n-1)(n-2)}{24} = 126$

$(n+1)n(n-1)(n-2) = 3024 = 9 \cdot 8 \cdot 7 \cdot 6$

Thus $n+1=9\Rightarrow n=8$.

Now ellipse $\frac{x^2}{16} + \frac{y^2}{n} = 1$

$\frac{x^2}{16} + \frac{y^2}{8} = 1$

Here $a^2 = 16$ and $b^2 = 8$ (since $16 > 8$).

Eccentricity:

$$\begin{gathered} e=\sqrt{1-\frac{b^2}{a^2}} \\ =\sqrt{1-\frac{8}{16}} \\ =\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}} \end{gathered}$$

$$\begin{gathered} Required Solution = 8^3\times 4-1+8^3\times 3+8^3\times 2 \\ =8^3(4+3+2)-1 \\ =8^3\left(9\right)-1 = 4607 \end{gathered}$$

$$\begin{gathered} \text{let the 5 digit number be abcde} \\ \text{Here}, a \ge 5, a+e \le 8 \\ \text{Also, not all b,c,d,e are} 0, if a=5 \\ \text{Now, if a=5 then} e\in \left\{3,2,1,0\right\} \\ \text{then}, 8^3.4-1\text{ ways}. \\ \text{If a=6, then e∈}\left\{2,1,0\right\} \\ \mathrm{then}, 8^3.3 \text{ways} \\ \text{if a=7, then e∈}\left\{1,0\right\} \\ {\text{then 8}}^3.2 \text{ways} \\ \text{Total ways} =\left(8^3\times 4-1\right)+8^3\times 3+8^3\times 2 \\ =4607 \end{gathered}$$

4 boys and 3 girls.
All girls together \(\Rightarrow G_1 G_2 G_3\)

Required number of ways= total number of ways in all girls and all boys are sit together$-$along this condition \(B_1\) and \(B_2\) are adjacent

$$\begin{gathered} =4!3!2!-2!3!3!2! \\ =144 \end{gathered}$$

number of numbers formed using digits $\left(1111133\right)=\frac{7!}{5!2!}=21$

number of numbers formed using digits $\left(1111223\right)=\frac{7!}{4!2!}=105$

number of numbers formed using digits $\left(1112222\right)=\frac{7!}{4!3!}=35$

Total numbers $=161$

Number of ways to choose 10 questions = Total ways $-$ number of ways choosing at most 3 questions from part $A$

${}^{13}{C}_{10}-{}^6{C}_3=286-20=266$

${40}^n=2^{3n}\times 5^n$

Now, highest power of 2 which divides \(60!\) is

$E_2\left(60!\right)=\left[\frac{60}{2}\right]+\left[\frac{60}{2^2}\right]+\left[\frac{60}{2^3}\right]+\left[\frac{60}{2^4}\right]+\left[\frac{60}{2^5}\right]$

$=30+15+7+3+1=56$

Now, highest power of 5 which divides \(60!\) is

$E_5\left(60!\right)=\left[\frac{60}{5}\right]+\left[\frac{60}{5^2}\right]$

\(=12+2=14\)

${40}^n={\left(2^3\right)}^n\times 5^n={\left(2^3\times 5\right)}^n$

$60!=2^{56}\times 5^{14}\dots =2^{14}\cdot {\left(2^3\cdot 5\right)}^{14}$

Maximum value of \(n\) is \(14\).

Total ways to partition \(5\) into \(4\) parts are :
\((5,0,0,0) \Rightarrow 1\) way
\((4,1,0,0) \Rightarrow \frac{5!}{4!}=5\) ways
\((3,2,0,0) \Rightarrow \frac{5!}{3!2!}=10\) ways
\((2,2,0,1) \Rightarrow \frac{5!}{2!2!2!}=15\) ways
\((2,1,1,1) \Rightarrow \frac{5!}{2!(1!)^3 3!}=10\) ways
\((3,1,1,0) \Rightarrow \frac{5!}{3!2!}=10\) ways
Total \(\Rightarrow 1+5+10+15+10+10=51\) ways

$$\begin{gathered} \mathrm{Case} 1: 1,1,1,1,1,3,3 \to \mathrm{Total} 7 \mathrm{possible} \mathrm{numbers} \\ \mathrm{then} \mathrm{total} \mathrm{number} \mathrm{of} \mathrm{arrangments} \mathrm{are} \frac{7!}{5!.2!}=21 \\ \mathrm{Case} 2: 1,1,1,1,2,2,3 \to \mathrm{Total} 7 \mathrm{possible} \mathrm{numbers} \\ \mathrm{total} \mathrm{arrangement} \mathrm{will} \mathrm{be}= \frac{7!}{4!.2!} =105 \\ \mathrm{Case} 3: 1,1,1,2,2,2,2 \to \mathrm{Total} 7 \mathrm{possible} \mathrm{numbers} \\ \mathrm{total} \mathrm{arrangement} \mathrm{for} \mathrm{this} \mathrm{case} =\frac{7!}{4!.3!}=35 \\ \mathrm{So} \mathrm{total} \mathrm{number} \mathrm{of} \mathrm{required} \mathrm{arrangments} = 21+105+35 = 161 \end{gathered}$$

$$\begin{gathered} Required Solution = 8^3\times 4-1+8^3\times 3+8^3\times 2 \\ =8^3(4+3+2)-1 \\ =8^3\left(9\right)-1 = 4607 \end{gathered}$$

\({ }^{n-1} C_r=\left(k^2-8\right)^n C_{r+1}\)

\(\underbrace{r+1 \geq 0, r \geq 0}_{r \geq 0}\)

\(\frac{{ }^n C_{r+1}}{{ }^n C_{r+1}}=k^2-8\)

\(\frac{r+1}{n}=k^2-8\)

\(\Rightarrow k^2-8>0\)

\((k-2 \sqrt{2})(k+2 \sqrt{2})>0\)

\(k \in(-\infty,-2 \sqrt{2}) \cup(2 \sqrt{2}, \infty) .........(1)\)
\(\therefore n \geq r+1, \frac{r+1}{n} \leq 1\)

\(\Rightarrow k^2-8 \leq 1\)

\(k^2-9 \leq 0\)

\(-3 \leq k \leq 3 ........(2)\)

From equation (1) and (2) we get

\(\mathrm{k} \in[-3,-2 \sqrt{2}) \cup(2 \sqrt{2}, 3]\)

Total number of ways such that at least 3 digits will not come in its position. =Derangement of 3 digits +Derangement of 4 digits +Derangement of 5 digits

$$\begin{gathered} ={}^{5}C_3\times 3!\times \left(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}\right) \\ +{}^{5}C_4\times 4!\times \left(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}\right) \\ +{}^{5}C_5\times 5!\times \left(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}\right) \\ =10\times 2+5\times 9+44=109 \end{gathered}$$

Given word, "KANPUR"

Words starting with \(A = 5!=120\)

Words starting with \(K = 5!=120\)

Words starting with \(N = 5!=120\)

Words starting with \(PA = 4!=24\)

Words starting with \(PK = 4!=24\)

Words starting with \(PN = 4!=24\)

Words starting with \(PRA = 3!=6\)

Next word is \(PRKANU = 1\)

FInal word is \(PRKAUN = 1\)

Total sum \(= 440\)

Hence, \(PRKAUN\) is the ${440}^{th}$ word

Let oranges are identical then

$x_1+x_2+x_3+x_4=16$ and $x_1,x_2,x_3,x_4\ge 1$

or $x_1^{\text{'}}+x_2^{\text{'}}+x_3^{\text{'}}+x_4^{\text{'}}=12$

so total number of solutions are

${=}^{12+3}{C}_3{=}^{15}{C}_3=455$

\(\sum_{r=1}^{30} \frac{r^2{ }^{30} c_r \cdot{ }^{30} C_r}{{ }^{30} C_{r-1}}=\sum_{r=1}^{30} r^2\left(\frac{30-r+1}{r}\right)\cdot{ }^{30} C_r\)

\(=\sum_{r=1}^{30} r(31-r) \cdot{ }^{30} C_r\)

\(=31 \sum_{r=1}^{30} r^{30} c_r-\sum_{r=1}^{30} r^2 \cdot c_r\)

\(=31\left(30 \cdot 2^{29}\right)-30(31) \cdot 2^{28}\)

\(=31 \times 15 \cdot 2^{29}\)

$$\begin{gathered} \text{ Total number of ways } \\ ={}^{12}C_1+{}^{12}C_2+{}^{12}C_3+{}^{12}C_4 \\ =12+66+220+495 \\ = 793 \end{gathered}$$

Let oranges are identical then

$x_1+x_2+x_3+x_4=16$ and $x_1,x_2,x_3,x_4\ge 1$

or $x_1^{\text{'}}+x_2^{\text{'}}+x_3^{\text{'}}+x_4^{\text{'}}=12$

so total number of solutions are

${=}^{12+3}{C}_3{=}^{15}{C}_3=455$

Required number of ways=

\begin{aligned}
& 8!-6!\times 3! \\
& =36,000
\end{aligned}

No of triangles =¹² C₃ –⁵ C₃
= 220–10
= 210

The middle letter is fixed as 'M'. The remaining 4 letters must be chosen from the alphabet set, such that they are distinct and in increasing order. Since order matters (increasing order), the number of ways to select and arrange the remaining letters would follow a certain combinatorial pattern.

\({ }^{12} C_2 \times{ }^{13} C_2\)

Since \(4\) different books are to be put into \(3\) different bags and no bag is empty,

the distribution must be \((2,1,1)\)

Choose the \(2\) books for the bag having \(2\) books: \({}^{4}C_2=6\)

Choose the bag which will contain these \(2\) books: \(3\) ways

Arrange the remaining \(2\) books in the remaining \(2\) bags: \(2!\) ways

Total number of ways \(={}^{4}C_2\cdot3\cdot2!=6\cdot3\cdot2=36\)

Here, word DAUGHTER contains 3 vowels and 5 consonants

Total number of letters = 8

then number of possible words with no restriction = 8!

Now, if we need to take all vowels together, now the possible number of words = $6!\times 3!$

Now, the words in which all vowels are not together

$$\begin{gathered} =8!-6!\times 3! \\ =6![8\times 7-3\times 2\times 1] \\ =720[56-6] \\ =36000 \end{gathered}$$

Given word, "KANPUR"

Words starting with \(A = 5!=120\)

Words starting with \(K = 5!=120\)

Words starting with \(N = 5!=120\)

Words starting with \(PA = 4!=24\)

Words starting with \(PK = 4!=24\)

Words starting with \(PN = 4!=24\)

Words starting with \(PRA = 3!=6\)

Next word is \(PRKANU = 1\)

Final word is \(PRKAUN = 1\)

Total sum \(= 440\)

Hence, \(PRKAUN\) is the ${440}^{th}$ word

$$\begin{gathered} \text{ Total number of ways } \\ ={}^{12}C_1+{}^{12}C_2+{}^{12}C_3+{}^{12}C_4 \\ =12+66+220+495 \\ = 793 \end{gathered}$$

$$\begin{gathered} {}^n{C}_{r-1}=28 \Rightarrow {}^8{C}_2=28 \\ {}^n{C}_r=56 \Rightarrow {}^8{C}_3=56 \\ {}^n{C}_{r+1}=70 \Rightarrow {}^8{C}_4=70 \\ On comparing, we get n=8 \& r =3 \\ Now, A(4\cos t, 4\sin t), B(2\sin t, -2\cos t), C(1,0) \\ Centroid of ∆ABC, \\ x = \frac{4\cos t+2\sin t+1}{3}\Rightarrow 3x-1=4\cos t+2\sin t \\ y = \frac{4\sin t-2\cos t+0}{3}\Rightarrow 3y =4\sin t-2\cos t \\ Now, {(3x-1)}^2+{\left(3y\right)}^2=\alpha \\ {(4\cos t+2\sin t)}^2+{(4\sin t-2\cos t)}^2=\alpha \\ 16{\cos }^{2}t+4{\sin }^{2}t+16\sin t.\cos t+16{\sin }^{2}t+4{\cos }^{2}t-16\sin t. \cos t=\alpha \\ \alpha =20 \end{gathered}$$

Number of triangles: $p = \binom{n}{3}$
Number of quadrilaterals: $q = \binom{n}{4}$

Using the identity $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}$: $p + q = \binom{n}{3} + \binom{n}{4} = \binom{n+1}{4} = 126$

$\frac{(n+1)n(n-1)(n-2)}{24} = 126$ $(n+1)n(n-1)(n-2) = 3024 = 9 \cdot 8 \cdot 7 \cdot 6$

Thus $n+1 = 9 \Rightarrow \boxed{n = 8}$.

For the ellipse $\frac{x^2}{16} + \frac{y^2}{n} = 1$: $\frac{x^2}{16} + \frac{y^2}{8} = 1$

Here $a^2 = 16$ and $b^2 = 8$ (since $16 > 8$).

Eccentricity: $e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{8}{16}}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}$

\(\alpha=\frac{(4!)!}{(4!)^{3!}},\ \beta=\frac{(5!)!}{(5!)^{4!}}\)

\(\alpha=\frac{24!}{(4!)^6},\ \beta=\frac{120!}{(5!)^{24}}\)

Let \(24\) distinct objects are divided into \(6\) groups of \(4\) objects in each group.

No. of ways of formation of group \(=\frac{24!}{(4!)^6\cdot 6!}\in N\)

Similarly,

Let \(120\) distinct objects are divided into \(24\) groups of \(5\) objects in each group.

No. of ways of formation of groups \(=\frac{120!}{(5!)^{24}\cdot 24!}\in N\)

\(\alpha=\frac{(4!)!}{(4!)^{3!}},\ \beta=\frac{(5!)!}{(5!)^{4!}}\)

\(\alpha=\frac{24!}{(4!)^6},\ \beta=\frac{120!}{(5!)^{24}}\)

Let \(24\) distinct objects are divided into \(6\) groups of \(4\) objects in each group.

No. of ways of formation of group \(=\frac{24!}{(4!)^6\cdot 6!}\in N\)

Similarly,

Let \(120\) distinct objects are divided into \(24\) groups of \(5\) objects in each group.

No. of ways of formation of groups \(=\frac{120!}{(5!)^{24}\cdot 24!}\in N\)

MATHS has only 5 letters, so in a 6-letter word at least one letter has to repeat.

Case 1: Exactly one letter is repeated twice
= ${}^{5}C_1·\frac{6!}{2!}=5\times 360=1800$

Case 2: Exactly two letters are repeated twice
=${}^{5}C_2·{}^{3}C_2\frac{6!}{2!2!}=5400$

Case 3: Exactly 3 letters are repeated twice =${}^{5}C_3\frac{6!}{2!2!2!}=900$

Hence the required solution = $1800+5400+900=8100$

\(abcde\)

\(a\geq 5\)

\(a+e\leq 8\)

Not all \(b,c,d,e\) are zero if \(a=5\).

\(\Rightarrow\) (i) \(a=5\)

\(5---\frac{e}{8}\)

\(8^3\cdot 4-1\) ways

\(\Rightarrow e\in\{3,2,1,0\}\)

\(\Rightarrow\) (ii) \(a=6\)

\(6---\underline{e}\Rightarrow e\in\{2,1,0\}\)

\(8^3\cdot 3\) ways

\(\Rightarrow\) (iii) \(a=7\)

\(7---\frac{e}{2}\)

\(\Rightarrow 8^3\cdot 2\) ways

\(\Rightarrow e\in\{1,0\}\)

\(\Rightarrow\) Total ways

\(\Rightarrow 8^3(2+3+4)-1=4608-1=4607\)

$$\begin{gathered} \text{Given word is GARDEN} \\ \text{Total possible words=n(S)}= 6! \\ \text{ when} A \& E\text{ are in order} \\ \text{Favourable case} ={}^6\mathrm{C}_2·4! \\ \text{Probabllity }\left(P\right) =\frac{ \text{favourable case}}{ \text{Total case}} \\ p=\frac{\left(15\right)4!}{\left(30\right)4!} \\ \text{Probablity when vowels are not in order} \\ =1-\frac{1}{2}=\frac{1}{2} \end{gathered}$$

\(\begin{aligned}
&\text { Correct option is (1) } \mathbf{1 6 8 0}\\
&\begin{aligned}
& \left\{2,2^2, 2^3 \ldots \ldots .^9\right\} \\
& \Rightarrow \frac{9!}{3!3!3!3!} \times 3!=\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4}{6 \times 6} \\
& =1680
\end{aligned}
\end{aligned}\)

Given word, "KANPUR"

Words starting with \(A = 5!=120\)

Words starting with \(K = 5!=120\)

Words starting with \(N = 5!=120\)

Words starting with \(PA = 4!=24\)

Words starting with \(PK = 4!=24\)

Words starting with \(PN = 4!=24\)

Words starting with \(PRA = 3!=6\)

Next word is \(PRKANU = 1\)

FInal word is \(PRKAUN = 1\)

Total sum \(= 440\)

Hence, \(PRKAUN\) is the ${440}^{th}$ word

$$\begin{gathered} {}^n{C}_{r-1}=28 \Rightarrow {}^8{C}_2=28 \\ {}^n{C}_r=56 \Rightarrow {}^8{C}_3=56 \\ {}^n{C}_{r+1}=70 \Rightarrow {}^8{C}_4=70 \\ On comparing, we get n=8 \& r =3 \\ Now, A(4\cos t, 4\sin t), B(2\sin t, -2\cos t), C(1,0) \\ Centroid of ∆ABC, \\ x = \frac{4\cos t+2\sin t+1}{3}\Rightarrow 3x-1=4\cos t+2\sin t \\ y = \frac{4\sin t-2\cos t+0}{3}\Rightarrow 3y =4\sin t-2\cos t \\ Now, {(3x-1)}^2+{\left(3y\right)}^2=\alpha \\ {(4\cos t+2\sin t)}^2+{(4\sin t-2\cos t)}^2=\alpha \\ 16{\cos }^{2}t+4{\sin }^{2}t+16\sin t.\cos t+16{\sin }^{2}t+4{\cos }^{2}t-16\sin t. \cos t=\alpha \\ \alpha =20 \end{gathered}$$

3 Shelf empty : \((8,0,0,0) \rightarrow 1\) way

2 shelf empty: \(\left.\begin{array}{l}(7,1,0,0) \\ (6,2,0,0) \\ (5,3,0,0) \\ (4,4,0,0)\end{array}\right] \rightarrow 4\) ways

1 shelf empty : \(\left.\begin{array}{ll}(6,1,1,0) & (3,3,2,0) \\ (5,2,1,0) & (4,2,2,0) \\ (4,3,1,0) & \end{array}\right] \rightarrow 5\) ways

0 Shelf empty \(\left.: \begin{array}{ll}(1,2,3,2) & (5,1,1,1) \\ (2,2,2,2) & \\ (3,3,1,1) & \\ (4,2,1,1) & \end{array}\right] \rightarrow 5\) ways

Total = \(15\) ways

3 Shelf empty : \((8,0,0,0) \rightarrow 1\) way

2 shelf empty: \(\left.\begin{array}{l}(7,1,0,0) \\ (6,2,0,0) \\ (5,3,0,0) \\ (4,4,0,0)\end{array}\right] \rightarrow 4\) ways

1 shelf empty : \(\left.\begin{array}{ll}(6,1,1,0) & (3,3,2,0) \\ (5,2,1,0) & (4,2,2,0) \\ (4,3,1,0) & \end{array}\right] \rightarrow 5\) ways

0 Shelf empty \(\left.: \begin{array}{ll}(1,2,3,2) & (5,1,1,1) \\ (2,2,2,2) & \\ (3,3,1,1) & \\ (4,2,1,1) & \end{array}\right] \rightarrow 5\) ways

Total = \(15\) ways

Available exit floors are \(3,4,5,6,7,8,9,10\)

So, number of possible floors \(=8\)

Choose \(4\) persons out of \(9\) persons:

\({ }^9 C_4=126\)

Choose one floor for these \(4\) persons and a different floor for the remaining \(5\) persons:

\(8 \times 7=56\)

Total number of ways \(={ }^9 C_4 \times 8 \times 7=126 \times 56=7056\)

$\text{INCONSEQUENTIAL}$ contains vowels $\text{I, E, O, A,U}$ and consonants $\text{N,C,S,Q,T,L}$

Total number of \(4\) letter words using two vowels and two consonants is

\(={ }^5 \mathrm{C}_2 \times{ }^6 \mathrm{C}_2 \times 4!\)

\(=3600\)