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atulkumar79 · Thu Jul 02, 2009 9:57 pm

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

OA C

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Robinmrtha · Thu Jul 02, 2009 10:47 pm

the sum of the digits can be five
in the following cases
when the digits of the number are from the following sets
1. (0,0,0,5)
2.(1,0,0,4)
3.(1,2,2,0)
4.(3,2,0,0)
5.(3,1,1,0)
6.(2,1,1,1)

from set one the no. of digits with sum 5 = 4!/3!=4
from set two the no. of digits with sum 5 = 4!/2!=12
from set three the no. of digits with sum 5 = 4!/2!=12
from set four the no. of digits with sum 5 = 4!/2!=12
from set five the no. of digits with sum 5 = 4!/2!=12
from set six the no. of digits with sum 5 = 4!/3!=4

Therefore, the answer is 4 +12+12+12+12+4=56

OA C

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getso · Thu Jul 02, 2009 11:30 pm

Hi,

Could you please explain permutation part. I could not understand.

Thanks,
Shobha

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Robinmrtha · Fri Jul 03, 2009 12:29 am

lets take an example...
in how many ways the alphabets of the word apple be arranged to form new words(may not have meaning)

apple has 5 alphabets
where there are 2 ps...
so even if we interchange p the word will remain same...
so before we take this into consideration the permutation will be 5!
But since p is repeated...we got to divide 5! by 2!
i.e. 5!/2!

I have done the same in above explanation...
in set one total digits is 4 but there are 3 zeros....
so i have divided 4!/3!
hope this helps....

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