(72)

Sum of first two digits

Sum of last two digits = $\alpha$

Case–I : $\alpha$ = 7

2 × 12 = 24 ways.

2 × 8 ways = 16 ways

2 × 4 ways = 8 ways

Total ways = 24 + 16 + 16 + 8 + 8 = 72

(514)

Number of numbers divisible by 2 = 450
Number of numbers divisible by 3 = 300
Number of numbers divisible by 7 = 128
Number of numbers divisible by 2 & 7 = 64
Number of numbers divisible by 3 & 7 = 43
Number of numbers divisible by 2 & 3 = 150
Number of numbers divisible by 2, 3 & 7 = 21

Total numbers = 450 + 300 – 150 – 64 – 43 + 21 = 514

(50400)

In the word ASSASSINATION, there are 6 vowels A A A I I O and consonants S S S S N N T.

$\text{Taking }\left(\mathrm{AAAIIO}\right)\text{ as a single unit, we get:}$

$\text{Number of ways to arrange ASSASSINATION such that all vowels occur together}$

$=\frac{8!}{4! 2!}\times \frac{6!}{3! 2!}=\frac{8!\times 6!}{4! {(2!)}^2 3!}=50,400$

(81)

Total number of 6-digit numbers formed by the digits 4, 5, 9 and divisible by 6 is when all the digits are same.

$444444\to \frac{6!}{6!}=1$

$\text{Taking two digits (4, 5), }555444\to \frac{5!}{3! 2!}=10$

$\text{For (4, 9), }999444\to \frac{5!}{3! 2!}=10$

$\text{Taking three digits: }459444\to \frac{5!}{3!}=20$

$559554\to \frac{5!}{4!}=5, 959994\to \frac{5!}{4!}=5$

$959454\to \frac{5!}{2! 2!}=30$

$\text{Total number}=20+5+5+30+10+10+1=81$

(60)

Even numbers are 2, 4, 2, 4 

Odd numbers are  1, 3, 1, 3, 1

$\overline{O} \overline{E} \overline{O} \overline{E} \overline{O} \overline{E} \overline{O} \overline{E} \overline{O}$

$\text{Even places can be filled with even numbers in }\frac{4!}{2! 2!}\text{ ways.}$

$\text{Similarly, odd places can be filled in }\frac{5!}{2! 3!}\text{ ways.}$

$\text{Required number of ways }=\frac{4!}{2! 2!}\times \frac{5!}{2! 3!}=60$

(120)

Let $x+y=5\lambda$

Possible cases are

$\begin{array}{ccccc}x& & y& & \text{Number of ways}\\ 5\lambda & & 5\lambda & & 20\\ 5\lambda +1& & 5\lambda +4& & 25\\ 5\lambda +2& & 5\lambda +3& & 25\\ 5\lambda +3& & 5\lambda +2& & 25\\ 5\lambda +4& & 5\lambda +1& & 25\end{array}$

$\text{Total number of ways }=20+25+25+25+25=120$

(1436)

$\text{Number of 5-digit numbers starting with digit 1 }=5\times 5\times 5\times 5=625$

$\text{No. of 5-digit numbers starting with digit 2 }=5^4=625$

$\text{No. of 5-digit numbers starting with 31 }=5\times 5\times 5=125$

$\text{No. of 5-digit numbers starting with 32 }=5\times 5\times 5=125$

$\text{No. of 5-digit numbers starting with 33 }=5\times 5\times 5=125$

$\text{No. of 5-digit numbers starting with 351 }=5\times 5=25$

$\text{No. of 5-digit numbers starting with 352 }=5\times 5=25$

$\text{No. of 5-digit numbers starting with 3531 }=5$

$\text{No. of 5-digit numbers starting with 3532 }=5$

$\text{Before 35337, there will be 4 numbers, so the rank of 35337 will be 1690.}$

$\text{So, in descending order serial number will be,}$ $3125-1690+1=1436$

(3000)

$\text{The total 4-digit numbers is }N$

$=9\times 10\times 10\times 10=9000$

$\text{The number }N\text{ should be divisible by 2 but not by 3.}$

$N=\left(\text{Number divisible by 2}\right)-\left(\text{Number divisible by 6}\right)$

$=\frac{9000}{2}-\frac{9000}{6}=4500-1500=3000$

(21)

$\text{We have to make 4-digit numbers using the digits }1,2,3,\text{ and }5.$

$\text{The unit digit of the 4-digit number will be }5.$

$\text{Now, the sum }(x+y+z)\text{ should be of the form }(3\lambda +1).$

$\text{Therefore, the possible cases are:}$

$(x,y,z)=(1,1,5),(1,1,2),(2,2,3),(2,3,5),(3,3,1),(5,5,3)$

$\text{So, total arrangements are:}$

$\text{For }(1,1,5)\to \frac{3!}{2!}=3 ; \text{For }(1,1,2)\to \frac{3!}{2!}=3$

$\text{For }(2,2,3)\to \frac{3!}{2!}=3 ; \text{For }(2,3,5)\to 3!=6$

$\text{For }(3,3,1)\to \frac{3!}{2!}=3 ; \text{For }(5,5,3)\to \frac{3!}{2!}=3$

$\text{So, total number of arrangements }=3+3+3+6+3+3=21$

(26)

We have, $\sum _{n=0}^{\infty }\frac{n^3(2n!)+(2n-1)(n!)}{(n!)\left(\right(2n)!)}=ae+\frac{b}{e}+c$

$\Rightarrow \sum _{n=0}^{\infty }\frac{1}{(n-3)!}+\sum _{n=0}^{\infty }\frac{3}{(n-2)!}+\sum _{n=0}^{\infty }\frac{1}{(n-1)!}+\sum _{n=0}^{\infty }\frac{1}{(2n-1)!}-\sum _{n=0}^{\infty }\frac{1}{\left(2n\right)!}=ae+\frac{b}{e}+c$

$\Rightarrow e+3e+e+\frac{1}{2}(e-\frac{{\displaystyle 1}}{{\displaystyle e}})-\frac{1}{2}(e+\frac{{\displaystyle 1}}{{\displaystyle e}})=ae+\frac{b}{e}+c$

$\Rightarrow 5e-\frac{1}{e}=ae+\frac{b}{e}+c$

$\text{Comparing, we get, }a=5, b=-1, c=0$

$a^2-b+c=25+1=26$