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I searched a lot to find some sort of guidance on how to approach solving this problem, but I couldn't find an answer anywhere. Read the book many times, still no luck The problem as follows:

Cecil has a 15-coin collection. Four coins are quarters, seven coins are dimes, three are nickels and one is a penny. For each scenario, calculate the total possible outcomes if Cecil randomly selects five coins.

Cecil selects no quarters
(11,5) 11! / 6! = 55440
Cecil selects at least 1 dime
(9,5) = 9! / 4! = 15120
Cecil selects an equal number of nickels and dimes
(6,5) = 6! / 1! = 720
The cent amount of Cecil’s selection is an even number
(7,5) = 7! /5! 2! = 2520

Are my answers to each scenario correct? If not, what am I missing? Thanks in advance for any answers.

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  • $\begingroup$ First of all: you should probably be working with combinations, not permutations, ie, $(11,5) = 11! / (5!*6!)$ $\endgroup$ – cpiegore Oct 12 '16 at 3:36
  • $\begingroup$ Second: the other questions will have several cases that will need to be considered separately. You will need to find the total possibilities for each case and add them. Example: for the second question add (# given 1 dime)+(# given 2 dimes)+(# given 3 dimes)+(# given 4 dimes)+(# given 5 dimes) $\endgroup$ – cpiegore Oct 12 '16 at 3:43
  • $\begingroup$ @cpiegore Thank you for the feedback! It looks like I confused the formulas. I corrected it now. $\endgroup$ – Crazy Computer Oct 12 '16 at 13:46
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You computed the number of ordered selections in which Cecil selects no quarters. However, we are interested in the number of subsets, so the selections are not ordered.

Cecil selects no quarters.

The number of ways Cecil can select five coins from the eleven coins that are not quarters is $$\binom{11}{5} = \frac{11!}{5!6!}$$

Cecil selects at least one dime.

Since there are seven dimes and eight other coins, one way to do this would be to add the number of ways Cecil could select one dime and four of the other eight coins, two dimes and three of the other eight coins, three dimes and two of the other eight coins, four dimes and one of the other eight coins, and five dimes.

$$\binom{7}{1}\binom{8}{4} + \binom{7}{2}\binom{8}{3} + \binom{7}{3}\binom{8}{2} + \binom{7}{4}\binom{8}{1} + \binom{7}{5}\binom{8}{0}$$

However, it is simpler to subtract the number of selections in which Cecil selects five coins that are not dimes from the total number of selections of five coins. Since seven of the fifteen available coins are dimes, this can be done in

$$\binom{15}{5} - \binom{8}{5}$$

ways.

Cecil selects an equal number of nickels and dimes.

There are seven dimes and three nickels. Since Cecil is selecting five of the fifteen coins, it is possible that he selects no nickels or dimes, one nickel and one dime, and two nickels and two dimes. He cannot select three of each since that would require the selection of at least six coins. If Cecil selects no nickels or dimes, he must select all five of the other coins. If he selects one nickel and one dime, he must select three of the other five coins. If he selects two nickels and two dimes, he must select one of the other five coins. Therefore, the number of ways in which Cecil can select an equal number of nickels and dimes is

$$\binom{7}{0}\binom{3}{0}\binom{5}{5} + \binom{7}{1}\binom{3}{1}\binom{5}{2} + \binom{7}{2}\binom{3}{2}\binom{5}{1}$$

The last question depends on knowing the values of the coins. They are $25$ cents for a quarter, $10$ cents for a dime, $5$ cents for a nickel, and $1$ cent for a penny.

Cecil selects an even number of cents.

For this to occur, the number of coins with odd cent values that are selected must be even, which means Cecil must select an odd number of dimes. Since there are seven dimes and he selects five coins, this means that he selects one dime and four of the other eight coins, three dimes and two of the other eight coins, or five dimes. Therefore, the number of selections of five coins which result in an even number of cents is

$$\binom{7}{1}\binom{8}{4} + \binom{7}{3}\binom{8}{2} + \binom{7}{5}\binom{8}{0}$$

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  • $\begingroup$ In the last question, the number of dimes does not need to be even. Examples: Cecil can select 1 dime and 4 quarters; 3 dimes and 2 quarters; 5 dimes; or 1 dime 3 nickels and 1 penny $\endgroup$ – cpiegore Oct 12 '16 at 14:36
  • $\begingroup$ Thank you very much for your help. You have explained this better than my professor, the TA, and the book. Hats off to you sir! I will attempt to do other questions based on what I learned from this questions. $\endgroup$ – Crazy Computer Oct 12 '16 at 14:50
  • $\begingroup$ @N. F. Taussig In fact the number of dimes must be odd because even+even=even and 5 is odd. $\endgroup$ – cpiegore Oct 12 '16 at 14:51
  • $\begingroup$ @cpiegore Quite correct. Thanks for alerting me to the error. $\endgroup$ – N. F. Taussig Oct 12 '16 at 20:05
  • $\begingroup$ @CrazyComputer Your praise may have been a bit too effusive. As cpiegore pointed out, the version you read earlier contained an error in the last problem. The number of dimes in the last problem must be odd rather than even, as I originally stated. I have revised my answer. $\endgroup$ – N. F. Taussig Oct 12 '16 at 20:08

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