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Being Positive!

by varun7nurav » Sun Sep 18, 2011 9:00 am
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

This sounded easy at first but latter got me all tangled up. Please help. Thanks!
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by GMATGuruNY » Sun Sep 18, 2011 9:19 am
I posted a solution here:

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by sl750 » Sun Sep 18, 2011 11:09 am
use zeros to fill the rest of the digits
Number of ways in which the sum of digits is 5 is
0005 = 4!/3! = 4
0014 = 4!/2! = 12
0023 = 4!/2! = 12
0122 = 4!/2! = 12
0113 = 4!/2! = 12
1112 = 4!/3! = 4

Total = 56 ways
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by varun7nurav » Sun Sep 18, 2011 11:39 am
Hi sl, ur method sounds easy. Nevertheless, gathering all the possibilities is another task.. Of the 6 combinations, i stand a chance to risk missing out one 1 or more.
How would we know that i need to come up with 6, and i may instead waste time making sure i got all...

Tq!
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by sl750 » Sun Sep 18, 2011 12:23 pm
Varun,
I don't think I can answer that one correctly; I guess it comes with practice. You know that you need a sum of 5, so you have to write down all possible combination's. There is a formula that you could use for such problems, but then again you will have to know when to use the formula

n+k-1Ck-1. Number of ways of distributing n identical things among k persons such that each person receives any number of things

Here n=5 (5 digit number), k=4(As we consider numbers less than 10000, the highest 4 digit number being 9999).
8C3 = 56
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by varun7nurav » Mon Sep 19, 2011 9:24 am
sl, Thanks a lot.. it cleared out the doubt i had.. your formula is very much similar to the solution given by GmatguruNY- 'seperator method'.

I guess, this formula applies to cases only when items are identical.

Cheers!
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