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

The Correct Option is D

Given: 

\[ {}^6C_m + 2({}^6C_{m+1}) + {}^6C_{m+2} > {}^8C_3 \]

\[ {}^7C_{m+1} + {}^7C_{m+2} > {}^8C_3 \]

\[ {}^8C_{m+2} > {}^8C_3 \]

∴ m = 2

Step 2: Using the relation

\[ {}^{n-1}P_3 : {}^nP_4 = 1 : 8 \]

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

∴ n = 8

Step 3:

\[ {}^nP_{m+1} + {}^{n+1}C_m = {}^8P_3 + {}^9C_2 \]

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

\[ = 372 \]

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