Question

Find $ nC_{21} $, if $ nC_{10} = nC_{12} $

• A. 1
• B. 21
• C. 22
• D. 2

Hint:

When solving for binomial coefficients, always use the symmetry property \( nC_r = nC_{n-r} \) to simplify the given equation.

Answer 1

The Correct Option is C

To solve for \( nC_{21} \) given that \( nC_{10} = nC_{12} \), we start by using the properties of combinations. Recall the identity:

\( nC_r = nC_{n-r} \)

This implies \( nC_{10} = nC_{12} \), we can express this as:

\( nC_{10} = nC_{n-10} \)

But given \( nC_{10} = nC_{12} \), it follows:

\( nC_{12} = nC_{n-12} \)

For these two expressions \( n-10 = 12 \) and \( n-12 = 10 \) to both be true, solve each:

1. From \( n-10 = 12 \):
2. From \( n-12 = 10 \):

Both equations give us \( n = 22 \).

Now, we need to find \( 22C_{21} \). Using the combination property:

\( nC_r = nC_{n-r} \)

This yields:

\( 22C_{21} = 22C_1 \)

It is well-known that:

\( 22C_1 = 22 \)

Thus, the value of \( nC_{21} \) is 22.

Answer 2

To solve for \( nC_{21} \) given that \( nC_{10} = nC_{12} \), we use the property of combinations: \( \binom{n}{r} = \binom{n}{n-r} \). 

This implies \( nC_{10} = nC_{12} \) leads to:

\(\binom{n}{10} = \binom{n}{12}\)

By the symmetry property: \(\binom{n}{r} = \binom{n}{n-r}\), we have \( n=10+12 \Rightarrow n=22 \).

The question asks for \( nC_{21} \) which using the property \( \binom{n}{n-1} = n \) gives us:

\(\binom{22}{21} = \binom{22}{1} = 22\).