Jee Main 2024 31 01 2024 Shift2 - Question11
Question 11
If for some $m, n ;{ }^{6} C _m+2\left({ }^{6} C _{m+1}\right)+{ }^{6} C _{m+2}>{ }^{8} C _3$ and ${ }^{n-1} P _3:{ }^{n} P _4=1: 8$, then ${ }^{n} P _{m+1}+{ }^{n+1} C _m$ is equal to
(1) 6756
(2) 7250
(3) 6223
(4) 6550
Show Answer
Answer: (1)
Solution:
${ }^{6} C _m+2\left({ }^{6} C _{m+1}\right)+{ }^{6} C _{m+2}>{ }^{8} C _3$
${ }^{7} C _{m+1}+{ }^{7} C _{m+2}>{ }^{8} C _3$
${ }^{8} C _{m+2}>{ }^{8} C _3$
$\therefore m=2$
and ${ }^{n-1} P _3:{ }^{n} P _4=1: 8$
$\frac{(n-1)(n-2)(n-3)}{n(n-1)(n-2)(n-3)}=\frac{1}{8}$
$\therefore n=8$
$\therefore{ }^{n} P _{m+1}+{ }^{n+1} C _m={ }^{8} P _5+{ }^{9} C _2$
$=8 \times 7 \times 6 \times 5 \times 4+\frac{9 \times 8}{2}$
$=6756$





