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Above 700 question!!!

by Ozlemg » Fri Jul 01, 2011 5:04 am
How many positive integers less than 10,000 are there in which sum of digits equals 5?
(a) 31
(b) 51
(c) 56
(d) 62
(e) 93
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by Frankenstein » Fri Jul 01, 2011 5:32 am
Hi,
Let abcd be the number. We need to find the number of non-negative integral solutions of a+b+c+d = 5
It is (5+4-1)C4-1 = 8C3 = 56
Hence, C

If x1+x2+..+xk = n, then the number of non-negative solutions is (n+k-1)C(k-1).
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by Ozlemg » Fri Jul 01, 2011 5:38 am
How did you get that it is a combination problem?
I could not! Does this kind of problems need "combination" approach?
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by Frankenstein » Fri Jul 01, 2011 5:58 am
Ozlemg wrote:How did you get that it is a combination problem?
I could not! Does this kind of problems need "combination" approach?
This is not a combination problem directly. But, by solving we end up with this formula. This will save time, else you need to consider cases:
1)count 1 digit numbers - only one(5)
2)count 2 digit numbers - five(14,23,32,41,50)
3)count 3 digit numbers ...
4)count 3 digit numbers ...
5)count 3 digit numbers ...
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by winniethepooh » Fri Jul 01, 2011 7:19 am
He is a human for sure!
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by GMATGuruNY » Fri Jul 01, 2011 9:08 am
Ozlemg wrote:How many positive integers less than 10,000 are there in which sum of digits equals 5?
(a) 31
(b) 51
(c) 56
(d) 62
(e) 93
Another approach is to use the separator method. Check here:

https://www.beatthegmat.com/experts-any- ... 82307.html
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by Frankenstein » Fri Jul 01, 2011 9:37 am
GMATGuruNY wrote: Another approach is to use the separator method. Check here:
https://www.beatthegmat.com/experts-any- ... 82307.html
Exactly. This separator method gives the formula (n+k-1)C(k-1).
Cheers!

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by baladon99 » Fri Jul 01, 2011 12:19 pm
the sum should be 5.maximum it can be a 4 digit number.

In the following cases the sum is 5.

1004 --> In 12 ways it can be represented ( 4!/2!)
1103---> 12 ways
2003---> 12 ways
2102---> 12 ways
1112----> 4 ways (4!/3!)
5000---> 4 ways

In total 56 ways :)
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