Question

If \[{}^n{C_{n - r}} + 3{}^n{C_{n - r + 1}} + 3{}^n{C_{n - r + 2}} + {}^n{C_{n - r + 3}} = {}^p{C_r}\] then prove that \[p = n + 3\].

Answer 1

Hint: First of all, split the terms and group the common suitable terms to use the formula \[{}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}}\] and reduce the terms. Continue this method until we get the last simplification. So, use this concept to reach the solution of the given problem.

Complete step-by-step answer:
The given expression is \[{}^n{C_{n - r}} + 3{}^n{C_{n - r + 1}} + 3{}^n{C_{n - r + 2}} + {}^n{C_{n - r + 3}} = {}^p{C_r}\]
\[
  {}^n{C_{n - r}} + {}^n{C_{n - r + 1}} + 2{}^n{C_{n - r + 1}} + 2{}^n{C_{n - r + 2}} + {}^n{C_{n - r + 2}} + {}^n{C_{n - r + 3}} = {}^p{C_r} \\
  \left( {{}^n{C_{n - r}} + {}^n{C_{n - r + 1}}} \right) + 2\left( {{}^n{C_{n - r + 1}} + {}^n{C_{n - r + 2}}} \right) + \left( {{}^n{C_{n - r + 2}} + {}^n{C_{n - r + 3}}} \right) = {}^p{C_r} \\
  \left( {{}^n{C_{n - r}} + {}^n{C_{\left( {n - r} \right) + 1}}} \right) + 2\left( {{}^n{C_{n - r + 1}} + {}^n{C_{\left( {n - r + 1} \right) + 1}}} \right) + \left( {{}^n{C_{n - r + 2}} + {}^n{C_{\left( {n - r + 2} \right) + 1}}} \right) = {}^p{C_r} \\
\]
We know that \[{}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}}\]
\[
  \left( {{}^{n + 1}{C_{n - r + 1}}} \right) + 2\left( {{}^{n + 1}{C_{n - r + 2}}} \right) + \left( {{}^{n + 1}{C_{n - r + 3}}} \right) = {}^p{C_r} \\
  {}^{n + 1}{C_{n - r + 1}} + {}^{n + 1}{C_{n - r + 2}} + {}^{n + 1}{C_{n - r + 2}} + {}^{n + 1}{C_{n - r + 3}} = {}^p{C_r} \\
  \left( {{}^{n + 1}{C_{n - r + 1}} + {}^{n + 1}{C_{\left( {n - r + 1} \right) + 1}}} \right) + \left( {{}^{n + 1}{C_{n - r + 2}} + {}^{n + 1}{C_{\left( {n - r + 2} \right) + 1}}} \right) = {}^p{C_r} \\
\]
By using the formula \[{}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}}\], again we have
\[
  \left( {{}^{n + 2}{C_{n - r + 2}}} \right) + \left( {{}^{n + 2}{C_{n - r + 3}}} \right) = {}^p{C_r} \\
  {}^{n + 2}{C_{n - r + 2}} + {}^{n + 2}{C_{\left( {n - r + 2} \right) + 1}} = {}^p{C_r} \\
\]
Again, using the formula \[{}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}}\], we get
\[{}^{n + 3}{C_{n - r + 3}} = {}^p{C_r}\]
We know that \[{}^n{C_r} = {}^n{C_{n - r}}\]
\[
  {}^{n + 3}{C_{\left( {n + 3} \right) - \left( {n - r + 3} \right)}} = {}^p{C_r} \\
  {}^{n + 3}{C_r} = {}^p{C_r} \\
  n + 3 = p{\text{ }}\left[ {\because {\text{If }}{}^n{C_r} = {}^p{C_r}{\text{ then }}n = p} \right] \\
  \therefore p = n + 3 \\
\]
Hence proved.

Note: In combinations, if \[{}^n{C_r} = {}^p{C_r}{\text{ }}\] then \[n = p\]. In these types of questions use the binomial theorem formulae of combinations to split and simplify terms as per our requirement to reach the solution and hence to prove it.