Chapter 11.7, Problem 29E

To determine

To explain ${}_5{C}_4$ = ${}_5{C}_1$ and to find another example of two different combinations that are equal

Explanation of Solution

Given information:

The given combinations is ${}_5{C}_4$ = ${}_5{C}_1$

Formula used: ${}_n{C}_r=\frac{n!}{r!\times (n-r)!}$

Proof:

  $\begin{array}{l}{}_n{C}_r{=}_5{C}_4\\ {}_n{C}_r=\frac{n!}{r!\times (n-r)!}\\ {}_5{C}_4=\frac{5!}{4!\times (5-4)!}\\ {}_5{C}_4=\frac{5!}{4!\times 1!}=\frac{5\times 4}{4\times 1}\\ {}_5{C}_4=5\text{ [Simplify}\text{.]}\end{array}$

  $\begin{array}{l}{}_n{C}_r{=}_5{C}_1\\ {}_n{C}_r=\frac{n!}{r!\times (n-r)!}\\ {}_5{C}_1=\frac{5!}{1!\times (5-1)!}\\ {}_5{C}_1=\frac{5!}{1!\times 4!}=\frac{5\times 4\times 3\times 2\times 1}{1\times 4\times 3\times 2\times 1}\\ {}_5{C}_1=5\text{ [Simplify}\text{.]}\\ Hence,{}_5{\text{C}}_4{=}_5{C}_1\text{ [Proved]}\end{array}$

Another example : ${}_3{C}_2{=}_3{C}_1$