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January 31, 2015

January 31, 2015

Posted by **Zoe** on Saturday, February 21, 2009 at 9:08pm.

In how many possible ways can the selection be made so that the value of the coins is at least 25 cents?

I know that the total outcomes equals 120, but how do I find how many of these have the value of at least 25 cents?

- math -
**Zoe**, Saturday, February 21, 2009 at 10:35pmThe answer is 92. Anyone have idea how to solve this or why the answer is 92? Thanks in advance!

- math -
**Reiny**, Sunday, February 22, 2009 at 10:20amConsider the cases which do not add up to 25, they would be

DNN and NNN

DNN -- C(4,1)C(4,2) = 24

NNN -- C(4,3 = 4

so 28 cases are not allowed

leaving 120 - 28 = 92

another way:

You have to list the possible outcomes , then evaluate each one, finally add up the allowable cases

QQQ - C(4,1) = 4

QQD - C(2,2)C(4,1) = 4

QNN - C(2,1)C(4,2) = 12

QDD - .... = 12

QND - .... = 32

DDD - .... = 4

DDN - .... = 24

DNN gives us < 25 cents

NNN gives us < 25 cents.

the total of the above is 92

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