Question

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

Answer 1

Hint: 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 \[{}^{18}{C_{15}} + 2\left( {{}^{18}{C_{16}}} \right) + {}^{17}{C_{16}} + 1 = {}^n{C_3}\]
Which can be rewrite as
\[ \Rightarrow {}^{18}{C_{15}} + {}^{18}{C_{16}} + {}^{18}{C_{16}} + {}^{17}{C_{16}} + 1 = {}^n{C_3}\]
By using the formula, \[{}^n{C_{r + 1}} + {}^n{C_r} = {}^{n + 1}{C_{r+1}}\] we get
\[
   \Rightarrow \left( {{}^{18}{C_{15}} + {}^{18}{C_{16}}} \right) + {}^{18}{C_{16}} + {}^{17}{C_{16}} + 1 = {}^n{C_3} \\
   \Rightarrow {}^{19}{C_{16}} + {}^{18}{C_{16}} + {}^{17}{C_{16}} + 1 = {}^n{C_3}{\text{ }}\left[ {\because {}^n{C_{r + 1}} + {}^n{C_r} = {}^{n + 1}{C_{r+1}}} \right] \\
\]
We know that \[{}^{17}{C_{17}} = 1\]. So, replacing 1 by \[{}^{17}{C_{17}}\] we get
\[ \Rightarrow {}^{19}{C_{16}} + {}^{18}{C_{16}} + {}^{17}{C_{16}} + {}^{17}{C_{17}} = {}^n{C_3}\]
Again, using the formula \[{}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r+1}}\] we get
\[
   \Rightarrow {}^{19}{C_{16}} + {}^{18}{C_{16}} + \left( {{}^{17}{C_{16}} + {}^{17}{C_{17}}} \right) = {}^n{C_3} \\
   \Rightarrow {}^{19}{C_{16}} + {}^{18}{C_{16}} + {}^{18}{C_{17}} = {}^n{C_3}{\text{ }}\left[ {\because {}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r+!}}} \right] \\
\]
By using the formula, \[{}^n{C_{r + 1}} + {}^n{C_r} = {}^{n + 1}{C_{r+1}}\] we get
\[
   \Rightarrow {}^{19}{C_{16}} + \left( {{}^{18}{C_{16}} + {}^{18}{C_{17}}} \right) = {}^n{C_3} \\
   \Rightarrow {}^{19}{C_{16}} + {}^{19}{C_{17}} = {}^n{C_3}{\text{ }}\left[ {\because {}^n{C_{r + 1}} + {}^n{C_r} = {}^{n + 1}{C_{r+1}}} \right] \\
   \Rightarrow {}^{20}{C_{17}} = {}^n{C_3}{\text{ }}\left[ {\because {}^n{C_{r + 1}} + {}^n{C_r} = {}^{n + 1}{C_{r+1}}} \right] \\
   \Rightarrow {}^{20}{C_3} = {}^n{C_3}{\text{ }}\left[ {\because {}^n{C_r} = {}^n{C_{n - r}}} \right] \\
\]
We know that if \[{}^n{C_r} = {}^s{C_r}\] then \[n = s\]. So, we have
\[\therefore n = 20\]
Thus, the correct option is B. 20

Note: Here we have used the formula of combinations as follows:
           1. \[{}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_r}\]
           2. If \[{}^n{C_r} = {}^s{C_r}\] then \[n = s\]
           3. \[{}^n{C_r} = {}^n{C_{n - r}}\]