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How many ways can an IRS auditor select 3 of 11 tax returns for an audit? There are 8 members on a board of directors. If they...Asked by Ashley
How many ways can an IRS auditor select 3 of 11 tax returns for an audit?
There are 8 members on a board of directors. If they must form a subcommittee of 3 members, how many different subcommittees are possible?
License plates are made using 3 letters followed by 3 digits. How many plates can be made if repetition of letters and digits is allowed.
At the first tri-city meeting, there are 8 people from town A, 7 people from town B, and 5 people from town C. If a council consisting of 5 people is randoml selected, find the probabilit that 3 are from town A and 2 are from town B.
There are 8 members on a board of directors. If they must form a subcommittee of 3 members, how many different subcommittees are possible?
License plates are made using 3 letters followed by 3 digits. How many plates can be made if repetition of letters and digits is allowed.
At the first tri-city meeting, there are 8 people from town A, 7 people from town B, and 5 people from town C. If a council consisting of 5 people is randoml selected, find the probabilit that 3 are from town A and 2 are from town B.
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Answered by
MathMate
The auditor chooses the first return out of 11, that makes 11 choices. For the second, she is left with 10, so there are 11*10 ways. For the third, she has 9 returns left from which to choose, so there are 11*10*9=990 ways to choosing the first 3 returns.
However, out of the 990 ways to choose the returns, each combination (say A,B,C) would have been repeated in 6 different orders, like ABC, BAC, CAB, BCA, ACB, CBA. The number of repetitions can be calculated by 6=3!, where three is the number of returns chosen.
Thus the number of ways she can choose 3 returns would be
11*10*9/3!=990/6=165.
It's your turn to choose the sub-committees and try the rest of the exercises.
If you have difficulties, post what you have or what the difficulties are.
However, out of the 990 ways to choose the returns, each combination (say A,B,C) would have been repeated in 6 different orders, like ABC, BAC, CAB, BCA, ACB, CBA. The number of repetitions can be calculated by 6=3!, where three is the number of returns chosen.
Thus the number of ways she can choose 3 returns would be
11*10*9/3!=990/6=165.
It's your turn to choose the sub-committees and try the rest of the exercises.
If you have difficulties, post what you have or what the difficulties are.
Answered by
bluestar
i need help me on math
like 2+2=
like 2+2=