6.Permutation and Combination
hard

If for some $\mathrm{m}, \mathrm{n} ;{ }^6 \mathrm{C}_{\mathrm{m}}+2\left({ }^6 \mathrm{C}_{\mathrm{m}+1}\right)+{ }^6 \mathrm{C}_{\mathrm{m}+2}>{ }^8 \mathrm{C}_3$ and ${ }^{n-1} P_3:{ }^n P_4=1: 8$, then ${ }^n P_{m+1}+{ }^{n+1} C_m$ is equal to

A

$380$

B

$376$

C

$384$

D

$372$

(JEE MAIN-2024)

Solution

${ }^6 \mathrm{C}_{\mathrm{m}}+2\left({ }^6 \mathrm{C}_{\mathrm{m}+1}\right)+{ }^6 \mathrm{C}_{\mathrm{m}+2}>{ }^8 \mathrm{C}_3$

${ }^7 \mathrm{C}_{\mathrm{m}+1}+{ }^7 \mathrm{C}_{\mathrm{m}+2}>{ }^8 \mathrm{C}_3$

${ }^8 \mathrm{C}_{\mathrm{m}+2}>{ }^8 \mathrm{C}_3$

$\therefore \mathrm{m}=2$

$\text { And }{ }^{\mathrm{n}-1} \mathrm{P}_3:{ }^{\mathrm{n} P_4=1: 8}$

$\frac{(n-1)(n-2)(n-3)}{n(n-1)(n-2)(n-3)}=\frac{1}{8}$

$\therefore \mathrm{n}=8$

$\therefore{ }^{\mathrm{n}} \mathrm{P}_{\mathrm{m}+1}+{ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{m}}={ }^8 \mathrm{P}_3+{ }^9 \mathrm{C}_2$

$=8 \times 7 \times 6+\frac{9 \times 8}{2}$

$=372$

Standard 11
Mathematics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.