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Re: A men's basketball league assigns every player a two-digit number for [#permalink]
Since repetation is not allowed so according to simple counting method
for place 1 5 possibilities
for place 2 4 possibility
total 20
Answer is:A
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Re: A men's basketball league assigns every player a two-digit number for [#permalink]
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Bunuel wrote:
A men's basketball league assigns every player a two-digit number for the back of his jersey. If the league uses only the digits 1-5, what is the maximum number of players that can join the league such that no player has a number with a repeated digit (e.g. 22), and no two players have the same number?

A. 20
B. 21
C. 22
D. 24
E. 25


We need to determine how many two-digit numbers can be created from 5 digits (1 to 5 inclusive), and no digits can be repeated. Since order matters, we have a permutation. Thus, the number of ways to create two-digit numbers is 5P2 = 5!/(5 - 2)! = !5 x 4 = 20.

Answer: A
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Re: A men's basketball league assigns every player a two-digit number for [#permalink]
Bunuel wrote:
A men's basketball league assigns every player a two-digit number for the back of his jersey. If the league uses only the digits 1-5, what is the maximum number of players that can join the league such that no player has a number with a repeated digit (e.g. 22), and no two players have the same number?

A. 20
B. 21
C. 22
D. 24
E. 25


The question is simply asking how many unique combinations are there for a basketball player numbers- the rest of the information is superfluous

For the first number there is a pool of 5 numbers and for the second numbers there is a pool of 4 numbers
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Re: A men's basketball league assigns every player a two-digit number for [#permalink]
Why do maximum answer should not be 50
54-4
The highest two digit number without repeating digits between 1 to 5 will be 54
Repeating number with same digits Like 11,22,33 and 44 Will be subtracted so 54-4
will be equal to 50
Means that Team can have 50 players
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Re: A men's basketball league assigns every player a two-digit number for [#permalink]
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