Very tricky counting problem

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by faraz_jeddah » Mon Jul 22, 2013 3:41 pm
sana.noor wrote:using the inserting stick and seperator technique

(5+3)!/5!.3! = 56...C is the answer

is my approach to answer this question right?
where did you come up with that formula/technique?

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by vipulgoyal » Mon Jul 22, 2013 10:06 pm
Hi Brent,

how did you come up with

" We can extend this solution and conclude that the number of integers less than 1,000,000 in which the sum of the digits equals 8 will be 13C5"
OR
8c3 for less then 10000

though I got the explanation how 13c5 and 8c3 are yeilding desired answers, but how did we come up with 13c5 and 8c3

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by Brent@GMATPrepNow » Tue Jul 23, 2013 6:44 am
vipulgoyal wrote:Hi Brent,

how did you come up with

" We can extend this solution and conclude that the number of integers less than 1,000,000 in which the sum of the digits equals 8 will be 13C5"
OR
8c3 for less then 10000

though I got the explanation how 13c5 and 8c3 are yeilding desired answers, but how did we come up with 13c5 and 8c3
DISCLAIMER: This question type is likely beyond the scope of the GMAT

Okay, let's begin with 8c3 for less then 10000
Let's take eight Os: OOOOOOOO
Let's randomly choose 3 of these O's and replace them with lines.
For example: OO|O|O|O
By counting the O's between lines, we see that this scenario represents the number 2111
Similarly, O|OO|O|O represents the number 1211, O||OOO|O represents the number 1031, and |O||OOOO represents the number 0104 (104)
In each selection, five O's remain, so the sum of the digits will always be 5.
So, we can select 3 O's from 8 O's in 8C3 ways (56 ways)

IMPORTANT: We want a 1-, 2-, 3- or 4-digit number (i.e., less than 10,000) such that the digits add to 5.
This is accomplished in (5 + 4 - 1)C(4 - 1) ways (8C3 ways)


Now let's deal with: the number of integers less than 1,000,000 in which the sum of the digits equals 8 will be 13C5
We want a 1-, 2-, 3-, 4-, 5-, or 6-digit number (i.e., less than 1,000,000) such that the digits add to 8.
This is accomplished in (8 + 6 - 1)C(6 - 1) ways (13C5 ways)

Cheers,
Brent
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by tanvis1120 » Fri Jul 25, 2014 9:54 am
What is insert stick and separator technique?

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by Brent@GMATPrepNow » Fri Jul 25, 2014 10:01 am
tanvis1120 wrote:What is insert stick and separator technique?
See the above posts.

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by GMATinsight » Sat Jul 26, 2014 1:17 pm
Just in case someone wants to remember one formula for questions like these...

Whole Number (0,1,2,3,...) Solutions of an Equation in r vaiables
= (n+r-1)C(r-1)


Where,
n = Number to be divided among r variables
r = Number of variables

Example:

In how many ways can 8 Chocolates of similar type be distributed among three children where a child can have 0 to 8 chocolates?

We form an equation A+B+C = 8
where A,B and C are the chocolates given to first, second and third child respectively

This equation can be solved as follows for finding whole number of solution of A, B and C
n = 8
r = 3
i.e. Whole number solutions = (8+3-1)C(3-1) = 10C2 = 45
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