Solve the following

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 + nCn+1Cr+ 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 = nC

or 20CnC   [∵ 20C17 = 20C20−17  20C3]

⇒ n = 20

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