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A men's basketball league assigns every player

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by BTGmoderatorDC » Wed Feb 28, 2018 4:28 pm
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

Can some experts show me the best solution in this?

OA A
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by DavidG@VeritasPrep » Wed Feb 28, 2018 5:43 pm
lheiannie07 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

Can some experts show me the best solution in this?

OA A
If there were no restrictions, and you had 5 options for the first digit and 5 options for the second digit, there'd have been 5*5 = 25 unique numbers.
But we know we can't have repeated numbers, so now we have to eliminate 11, 22, 33, 44, 55 --> 5 numbers we have to eliminate.
25 - 5 = 20. The answer is A
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by Scott@TargetTestPrep » Mon Jun 10, 2019 6:25 pm
BTGmoderatorDC 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

Can some experts show me the best solution in this?

OA A
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.

Alternate Solution:

If repeated digits were allowed, there would be 25 possibilities since for each digit we would have 5 choices. Among these 25 possibilities, 5 of them are repeated digit numbers (which are 11, 22, 33, 44, and 55). Thus, without the repeated digits, there are 25 - 5 = 20 numbers possible.

Answer: A

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