Chapter 11.7, Problem 34E

To determine

To find more combinations of 99 of the coins than there are of 100 coins

Answer to Problem 34E

There are more combinations of 99 of the coins

Explanation of Solution

Given information:

The given information is ${}_{100}{C}_{100}$ and ${}_{100}{C}_{99}$

Calculation:

  $\begin{array}{l}{}_n{C}_r{=}_{100}{C}_{100}\\ {}_n{C}_r=\frac{n!}{r!\left(n-r\right)!}\text{ [Use combinations formula]}\\ {}_{100}{C}_{100}=\frac{100!}{100!\left(100-100\right)}\\ {}_{100}{C}_{100}=\frac{100!}{100!\times 0!}=\frac{100}{100}\\ {}_{100}{C}_{100}=1\text{ [Simplify}\text{.]}\end{array}$

  $\begin{array}{l}{}_n{C}_r{=}_{100}{C}_{99}\\ {}_n{C}_r=\frac{n!}{r!\left(n-r\right)!}\text{ [Use combinations formula]}\\ {}_{100}{C}_{99}=\frac{100!}{99!\left(100-99\right)!}\\ {}_{100}{C}_{99}=\frac{100!}{99!\times 1!}=\frac{100\times 99}{99!\times 1!}\text{ [Divide out common]}\\ {}_{100}{C}_{99}=100\text{ [Simplify}\text{.]}\end{array}$

So, there are more combinations of 99 of the coins