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by vipulgoyal » Mon Jul 29, 2013 3:49 am
How many 4 digit numbers can be formed by using the digits 0-9 so that it contains exactly 3 distinct digits?
(A)1944
(B)3240
(C)3850
(D)3888
(E)4216
D
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by GMATGuruNY » Mon Jul 29, 2013 3:58 am
How many 4 digit numbers can be formed by using the digits 0-9 so that it contains exactly 3 distinct digits?
(A)1944
(B)3240
(C)3850
(D)3888
(E)4216
For there to be exactly 3 distinct digits, EXACTLY 2 of the digits must be THE SAME.

Case 1: Tens digit and units digit the same:
Number of options for the thousands digit = 9. (Any digit 1-9)
Number of options for the hundreds digit = 9. (Any digit 0-9 not yet chosen)
Number of options for the tens digit = 8. (Any digit 0-9 not yet chosen)
Number of options for the units digit = 1. (Must be the same as the tens digit)
To combine the options above, we multiply:
9*9*8*1 = 648.

Other cases:
Of the 4 digits, any pair could compose the two that are the same.
Number of combinations of 2 that can be formed from 4 options = 4C2 = 6.

Since there are 6 pairs of digits that could be the same, and there are 648 options for each pair, we get:
Total ways = 6*648 = 3888.

The correct answer is D.
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by luckypiscian » Thu Aug 01, 2013 12:54 am
Could not understand your answer ... please explain further

if you choose 0 as tens (repeated) digit in Case 1, no. of combinations possible per selection shall be 2*3*2*1/2 = 6
whereas if you choose 0 as hundreds digit, no. of combinations possible per selection shall be 3*3*2*1/2 = 9
and if you choose all three non 0 digits, no. of combinations possible per selection shall be 4*3*2*1/2 = 12

don't we need to consider 3 different cases ?
6*(9C2) = 216
9*(9C2*2) = 648
12*(9C3*3) = 3024
total = 3024 + 648 + 216 = 3888
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by GMATGuruNY » Thu Aug 01, 2013 6:44 am
luckypiscian wrote:Could not understand your answer ... please explain further

don't we need to consider 3 different cases ?
Number of cases:
A pair of digits must be the same.
Number of pairs that can be chosen from the 4 digits = 4C2 = (4*3)/(2*1) = 6.

Options for each case:
In each case, we have the following options:
Number of options for the thousands place = 9. (Any digit but 0.)
Number of options for the digit that is the SAME as another digit. = 1. (Must be the same as the other digit.)
Number of options for the next digit selected = 9. (Any digit 0-9 but the digit used in the thousands place.)
Number of options for the last digit selected = 8. (8 digits left.)
Total options in each case = 9*1*9*8.

To combine all of these options, we multiply:
6*9*1*9*8 = 3888.
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