Question:
If ${ }^{18} C_{15}+2\left({ }^{18} C_{16}\right)+{ }^{17} C_{16}+1={ }^{n} C_{3}$, then $n=$ __________________
Solution:
Given ${ }^{18} C_{15}+2\left({ }^{18} C_{16}\right)+{ }^{17} C_{16}+1={ }^{n} C_{3}$
L.H.S 18C15 + 18C16 + 18C16 + 17C16 + 1
Since, nCr+1 + nCr = n+1Cr+1 and 1 = 17C17
L.H.S reduces to,
19C16 + 18C16 + 17C16 + 17C17
=19C16 + 18C16 + 18C17
= 19C16 + 19C17
i.e L.H.S = 20C17 and R.H.S = nC3 = L.H.S (given)
∴ 20C17 = nC3
or 20C3 = nC3 [∵ 20C17 = 20C20−17 = 20C3]
⇒ n = 20
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