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28 Dec 2007, 03:42
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Two similar questions.
A men's backetball league assigns every player a 2 digit number for the back of his jersey. If the league uses only the digits 1-5,
First Question :
what is the maximum number of players that can join the league such that no player has a number with a repeated digit(eg:- 33) and no 2 players have the same number ?
Second question :
what is the maximum number of players that can join the league such that no 2 players have the same number ?
Appreciate if you could emphasise the formula when you explain.
A men's backetball league assigns every player a 2 digit number for the back of his jersey. If the league uses only the digits 1-5,
First Question :
what is the maximum number of players that can join the league such that no player has a number with a repeated digit(eg:- 33) and no 2 players have the same number ?
Second question :
what is the maximum number of players that can join the league such that no 2 players have the same number ?
Appreciate if you could emphasise the formula when you explain.
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28 Dec 2007, 04:56
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1. N1=5C1*5C1-5C1=5*5-5=20
2. N2=5C1*5C1=5*5=25
2. N2=5C1*5C1=5*5=25
28 Dec 2007, 05:08
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1. We are creating 2 digit numbers using the numbers 1-5.
We can choose the first number from 5 choices. So options for the first number = 5C1 or 5. Then we have just as many choices for the second number so 5C1 or 5 again.
So two digit numbers made up of 5 choices = 5C1*5C1=25. Then we have to go back and subtract out the duplicates, and we can intuitively see there are 5 of those (11,22,33,44,55) so 25-5=20
2. This is the same problem except that we don't have to subtract the number of repeating numbers (11, 22, etc) so 5C1*5C1=25
We can choose the first number from 5 choices. So options for the first number = 5C1 or 5. Then we have just as many choices for the second number so 5C1 or 5 again.
So two digit numbers made up of 5 choices = 5C1*5C1=25. Then we have to go back and subtract out the duplicates, and we can intuitively see there are 5 of those (11,22,33,44,55) so 25-5=20
2. This is the same problem except that we don't have to subtract the number of repeating numbers (11, 22, etc) so 5C1*5C1=25
