Binomial Theorem Question 12
Question 12 - 2024 (31 Jan Shift 2)
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) 380
(2) 376
(3) 384
(4) 372
Show Answer
Answer (4)
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 ${ }^{n-1} P _3:{ }^{n} P _4=1: 8$
$\frac{(n-1)(n-2)(n-3)}{n(n-1)(n-2)(n-3)}=\frac{1}{8}$
$\therefore n=8$
$\therefore{ }^{n} P _{m+1}+{ }^{n+1} C _m={ }^{8} P _3+{ }^{9} C _2$
$=8 \times 7 \times 6+\frac{9 \times 8}{2}$
$=372$
