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Question: If \[{}^{18}{C_{15}} + 2\left( {{}^{18}{C_{16}}} \right) + {}^{17}{C_{16}} + 1 = {}^n{C_3}\], then \...

If 18C15+2(18C16)+17C16+1=nC3{}^{18}{C_{15}} + 2\left( {{}^{18}{C_{16}}} \right) + {}^{17}{C_{16}} + 1 = {}^n{C_3}, then nn is equal to
A. 19
B. 20
C. 18
D. 24

Explanation

Solution

In this question, we will proceed by grouping the terms common wherever we can use the formula in combinations to simply the value for the further steps. Use the same method until it gets simplified to the required value.

Complete step-by-step answer :
Given that 18C15+2(18C16)+17C16+1=nC3{}^{18}{C_{15}} + 2\left( {{}^{18}{C_{16}}} \right) + {}^{17}{C_{16}} + 1 = {}^n{C_3}
Which can be rewrite as
⇒18C15+18C16+18C16+17C16+1=nC3\Rightarrow {}^{18}{C_{15}} + {}^{18}{C_{16}} + {}^{18}{C_{16}} + {}^{17}{C_{16}} + 1 = {}^n{C_3}
By using the formula, nCr+1+nCr=n+1Cr+1{}^n{C_{r + 1}} + {}^n{C_r} = {}^{n + 1}{C_{r+1}} we get

⇒(18C15+18C16)+18C16+17C16+1=nC3 ⇒19C16+18C16+17C16+1=nC3 [∵nCr+1+nCr=n+1Cr+1]  \Rightarrow \left( {{}^{18}{C_{15}} + {}^{18}{C_{16}}} \right) + {}^{18}{C_{16}} + {}^{17}{C_{16}} + 1 = {}^n{C_3} \\\ \Rightarrow {}^{19}{C_{16}} + {}^{18}{C_{16}} + {}^{17}{C_{16}} + 1 = {}^n{C_3}{\text{ }}\left[ {\because {}^n{C_{r + 1}} + {}^n{C_r} = {}^{n + 1}{C_{r+1}}} \right] \\\

We know that 17C17=1{}^{17}{C_{17}} = 1. So, replacing 1 by 17C17{}^{17}{C_{17}} we get
⇒19C16+18C16+17C16+17C17=nC3\Rightarrow {}^{19}{C_{16}} + {}^{18}{C_{16}} + {}^{17}{C_{16}} + {}^{17}{C_{17}} = {}^n{C_3}
Again, using the formula nCr+nCr+1=n+1Cr+1{}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r+1}} we get

⇒19C16+18C16+(17C16+17C17)=nC3 ⇒19C16+18C16+18C17=nC3 [∵nCr+nCr+1=n+1Cr+!]  \Rightarrow {}^{19}{C_{16}} + {}^{18}{C_{16}} + \left( {{}^{17}{C_{16}} + {}^{17}{C_{17}}} \right) = {}^n{C_3} \\\ \Rightarrow {}^{19}{C_{16}} + {}^{18}{C_{16}} + {}^{18}{C_{17}} = {}^n{C_3}{\text{ }}\left[ {\because {}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r+!}}} \right] \\\

By using the formula, nCr+1+nCr=n+1Cr+1{}^n{C_{r + 1}} + {}^n{C_r} = {}^{n + 1}{C_{r+1}} we get

⇒19C16+(18C16+18C17)=nC3 ⇒19C16+19C17=nC3 [∵nCr+1+nCr=n+1Cr+1] ⇒20C17=nC3 [∵nCr+1+nCr=n+1Cr+1] ⇒20C3=nC3 [∵nCr=nCn−r]  \Rightarrow {}^{19}{C_{16}} + \left( {{}^{18}{C_{16}} + {}^{18}{C_{17}}} \right) = {}^n{C_3} \\\ \Rightarrow {}^{19}{C_{16}} + {}^{19}{C_{17}} = {}^n{C_3}{\text{ }}\left[ {\because {}^n{C_{r + 1}} + {}^n{C_r} = {}^{n + 1}{C_{r+1}}} \right] \\\ \Rightarrow {}^{20}{C_{17}} = {}^n{C_3}{\text{ }}\left[ {\because {}^n{C_{r + 1}} + {}^n{C_r} = {}^{n + 1}{C_{r+1}}} \right] \\\ \Rightarrow {}^{20}{C_3} = {}^n{C_3}{\text{ }}\left[ {\because {}^n{C_r} = {}^n{C_{n - r}}} \right] \\\

We know that if nCr=sCr{}^n{C_r} = {}^s{C_r} then n=sn = s. So, we have
∴n=20\therefore n = 20
Thus, the correct option is B. 20

Note : Here we have used the formula of combinations as follows:
1. nCr+nCr+1=n+1Cr{}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_r}
2. If nCr=sCr{}^n{C_r} = {}^s{C_r} then n=sn = s
3. nCr=nCn−r{}^n{C_r} = {}^n{C_{n - r}}