PS - Integers 2

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by neelgandham » Tue Dec 06, 2011 1:59 am
Hi Karthik,

The question has already been solved/discussed here https://www.beatthegmat.com/very-tricky- ... 25349.html
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by Anurag@Gurome » Tue Dec 06, 2011 2:00 am
karthikpandian19 wrote:How many positive integers less than 10,000 are there in which the sum of the digits equals 5?

(A) 31 (B) 51(C) 56(D) 62(E) 93
Sum of the digits is 5, so the numbers can be formed with the digits 0 to 5.

Numbers formed with digits 0, 0, 0 and 5 (0005) = 4!/3! = 4
Numbers formed with digits 0, 0, 1 and 4 (0041) = 4!/2! = 12
Numbers formed with digits 0, 0, 2 and 3 (0032) = 4!/2! = 12
Numbers formed with digits 0, 1, 1 and 3 (0113) = 4!/2! = 12
Numbers formed with digits 0, 1, 2 and 2 (0122) = 4!/2! = 12
Numbers formed with digits 1, 1, 1 and 2 (1112) = 4!/3! = 4

Therefore, number of positive integers less than 10,000 are there in which the sum of the digits equals 5 = (4 * 2) + (12 * 4) = 8 + 48 = 56

The correct answer is C.
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