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Using the numbers 1, 3, 5, 7 and 9, there are x 4-digit

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Using the numbers 1, 3, 5, 7 and 9, there are x 4-digit [#permalink] New post 18 Apr 2007, 06:14
Using the numbers 1, 3, 5, 7 and 9, there are x 4-digit numbers, in which at least 2 digits are the same. Therefore, x must be:

(A) 505
(B) 427
(C) 120
(D) 625
(E) 384
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 [#permalink] New post 18 Apr 2007, 06:34
we can do this if we can find out 2 things:
1) The number of 4-digit numbers where none of the digits are the same.
2) The total number of 4-digits numbers that we can form with the given numbers.

2) - 1) will give us the answer.

To get 1)
5*4*3*2 = 120

To get 2)
5*5*5*5 = 625

thta gives 505 .....(A)
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 [#permalink] New post 18 Apr 2007, 06:44
correct and nice explanation
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Re: combination [#permalink] New post 18 Apr 2007, 21:26
andrehaui wrote:
Using the numbers 1, 3, 5, 7 and 9, there are x 4-digit numbers, in which at least 2 digits are the same. Therefore, x must be:

(A) 505
(B) 427
(C) 120
(D) 625
(E) 384


There are 5 possible numbers.
Number of all possible 4 digit numbers = 5^4 = 625
Number of 4 digit numbers with distinct numbers = 5.4.3.2 = 120
Hence answer is 625 - 120 = 505

A
Re: combination   [#permalink] 18 Apr 2007, 21:26
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Using the numbers 1, 3, 5, 7 and 9, there are x 4-digit

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