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Worked Examples · Example 17

Q.If nC9=nC8{}^{n}C_9 = {}^{n}C_8, find nC17{}^{n}C_{17}.

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✓ Free question

By the symmetry nCa=nCb{}^{n}C_a={}^{n}C_b with a≠b⇒a+b=na\ne b \Rightarrow a+b=n, the equality gives n=9+8=17n = 9+8 = 17, so nC17=17C17=1{}^{n}C_{17} = {}^{17}C_{17} = 1.

1. Use the combination identity. If nCa=nCb{}^{n}C_a = {}^{n}C_b then either a=ba=b or a+b=na+b=n.

2. Apply it here. Since 9≠89 \ne 8, the second case must hold:

9+8=n  ⇒  n=179 + 8 = n \;\Rightarrow\; n = 17

3. Evaluate the required combination.

nC17=17C17=1{}^{n}C_{17} = {}^{17}C_{17} = 1

because there is exactly one way to choose all 1717 objects from 1717.

✓Final answer

nC17=1{}^{n}C_{17} = 1.

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