Chapter 9, Problem 16RE

Solution

To state: The possible total amount in Sean’s pocket if he has less than 50 cents in American coins in his pocket and no two coins of the same denomination.

The resultant amount is 16.

Given information:

Sean has less than 50 cents in American coins in his pocket and no two coins of the same denomination.

Explanation:

The possible coin in Sean’s pocket can be penny, nickel, dime, and quarter because he does not have two coins of the same denomination.

Now the possible number of ways to select zero or more coins out of these 4 coins will become:

The number of combination of 4 coins, taken zero at a time: $C_{4_0}$

The number of combination of 4 coins, taken one at a time: $C_{4_1}$

The number of combination of 4 coins, taken two at a time: $C_{4_2}$

The number of combination of 4 coins, taken three at a time: $C_{4_3}$

The number of combination of 4 coins, taken four at a time: $C_{4_4}$

Then the possible number of ways will become: $C_{4_0}+C_{4_1}+C_{4_2}+C_{4_3}+C_{4_4}$

  $\begin{array}{c}{C}_{4_0}+C_{4_1}+C_{4_2}+C_{4_3}+C_{4_4}=\frac{4!}{0!\left(4-0\right)!}+\frac{4!}{1!\left(4-1\right)!}+\frac{4!}{2!\left(4-2\right)!}+\frac{4!}{3!\left(4-3\right)!}+\frac{4!}{4!\left(4-4\right)!}\\ =\frac{4!}{0!4!}+\frac{4\cdot 3!}{1!3!}+\frac{4\cdot 3\cdot 2!}{2!2!}+\frac{4\cdot 3!}{3!1!}+\frac{4!}{4!1!}\\ =1+4+6+4+1\\ =16\end{array}$

Therefore, the total number of amount in Sean’s pocket will be equal to 16.