Solution:
C (20, 10) + 2 C (20, 9) + C (20, 8) = C (20, 10) + C (20, 9) + C (20, 9) + C (20, 8)
As we know that, C (n, r) + C (n, r - 1) = C (n + 1, r)
⇒ C (20, 10) + C (20, 9) = C (21, 10)
Similarly, C (20, 9) + C (20, 8) = C (21, 9)
⇒ C (20, 10) + 2 C (20, 9) + C (20, 8) = C (21, 10) + C (21, 9)
Again by using the rule, C (n, r) + C (n, r - 1) = C (n + 1, r) for the above equation, we get
⇒ C (20, 10) + 2 C (20, 9) + C (20, 8) = C (22, 10)
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