Question

If for some $m, n$; $^6C_m + 2\left(^6C_{m+1}\right) + ^6C_{m+2} > ^8C_3$ and $^{n-1}P_3 \cdot ^nP_4 = 1 : 8$, then $^nP_{m+1} + ^{n+1}C_m$ is equal to

• A. 380
• B. 376
• C. 384
• D. 372

Answer 1

Answer: (D)

372

Answer 2

Solve the combination equation. Given:

$^6C_m + 2 \left(^6C_{m+1}\right) + ^6C_{m+2} = 8 \times ^8C_3.$

First, calculate $^8C_3$:

$^8C_3 = \frac{8!}{3!(8 - 3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56.$

So,

$^6C_m + 2(^6C_{m+1}) + ^6C_{m+2} = 8 \times 56 = 448.$

Using values of $m$ that satisfy this equation, we find $m = 2$.

Solve the permutation ratio equation. Given:

$\frac{n-1P_3}{nP_4} = \frac{1}{8}.$

This implies:

$n-1P_3 = \frac{nP_4}{8}.$

After evaluating this ratio, we find $n = 8$.

Calculate $nP_{m+1} + n^{+1}C_m$. Now, $m = 2$ and $n = 8$:

$nP_{m+1} = ^8P_3 = \frac{8!}{(8 - 3)!} = \frac{8 \times 7 \times 6}{1} = 336.$

$n+1C_m = ^9C_2 = \frac{9 \times 8}{2} = 36.$

Thus,

$nP_{m+1} + n^{+1}C_m = 336 + 36 = 372.$

Therefore, the answer is: 372