Question

If ⁿC₅ + ⁿC₆ = ⁵¹C₆, then the value of n is equal to

• A. 49
• B. 50
• C. 45
• D. 46
• E. 51

Answer 1

The Correct Option is B

We are given the equation: \[ nC_5 + nC_6 = 51C_6 \] Using Pascal's identity: \[ nC_r + nC_{r+1} = (n+1)C_{r+1} \] we can apply this identity to simplify the left-hand side: \[ nC_5 + nC_6 = (n+1)C_6 \] Thus, we have the equation: \[ (n+1)C_6 = 51C_6 \] Since the binomial coefficients \( C_6 \) are equal, we can equate the numbers: \[ n+1 = 51 \] Solving for \( n \): \[ n = 50 \]

The correct option is (B) : \(50\)

Answer 2

The given equation is:

\[ \binom{n}{5} + \binom{n}{6} = \binom{51}{6} \]

We can use Pascal's identity, which states:

\[ \binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1} \]

Using this identity, we rewrite the left-hand side of the equation as:

\[ \binom{n}{5} + \binom{n}{6} = \binom{n+1}{6} \]

Thus, the equation becomes:

\[ \binom{n+1}{6} = \binom{51}{6} \]

Since the binomial coefficients are equal, we must have:

\[ n+1 = 51 \]

Therefore, the value of \(n\) is:

\[ n = 50 \]

Thus, the value of \(n\) is 50.