Three coins are selected from 10 coins: 4 dimes, 4 nickels, and 2 quarters.
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?
2 answers
The answer is 92. Anyone have idea how to solve this or why the answer is 92? Thanks in advance!
Consider 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
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
4 answers