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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}...

If nCn−r+3nCn−r+1+3nCn−r+2+nCn−r+3=pCr{}^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+3p = n + 3.

Explanation

Solution

Hint: First of all, split the terms and group the common suitable terms to use the formula nCr+nCr+1=n+1Cr+1{}^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 nCn−r+3nCn−r+1+3nCn−r+2+nCn−r+3=pCr{}^n{C_{n - r}} + 3{}^n{C_{n - r + 1}} + 3{}^n{C_{n - r + 2}} + {}^n{C_{n - r + 3}} = {}^p{C_r}

nCn−r+nCn−r+1+2nCn−r+1+2nCn−r+2+nCn−r+2+nCn−r+3=pCr (nCn−r+nCn−r+1)+2(nCn−r+1+nCn−r+2)+(nCn−r+2+nCn−r+3)=pCr (nCn−r+nC(n−r)+1)+2(nCn−r+1+nC(n−r+1)+1)+(nCn−r+2+nC(n−r+2)+1)=pCr  {}^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 nCr+nCr+1=n+1Cr+1{}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}}

(n+1Cn−r+1)+2(n+1Cn−r+2)+(n+1Cn−r+3)=pCr n+1Cn−r+1+n+1Cn−r+2+n+1Cn−r+2+n+1Cn−r+3=pCr (n+1Cn−r+1+n+1C(n−r+1)+1)+(n+1Cn−r+2+n+1C(n−r+2)+1)=pCr  \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 nCr+nCr+1=n+1Cr+1{}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}}, again we have

(n+2Cn−r+2)+(n+2Cn−r+3)=pCr n+2Cn−r+2+n+2C(n−r+2)+1=pCr  \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 nCr+nCr+1=n+1Cr+1{}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}}, we get
n+3Cn−r+3=pCr{}^{n + 3}{C_{n - r + 3}} = {}^p{C_r}
We know that nCr=nCn−r{}^n{C_r} = {}^n{C_{n - r}}

n+3C(n+3)−(n−r+3)=pCr n+3Cr=pCr n+3=p [∵If nCr=pCr then n=p] ∴p=n+3  {}^{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 nCr=pCr {}^n{C_r} = {}^p{C_r}{\text{ }} then n=pn = 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.