where did you come up with that formula/technique?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?
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Very tricky counting problem
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faraz_jeddah
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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
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
GMAT/MBA Expert
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Brent@GMATPrepNow
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DISCLAIMER: This question type is likely beyond the scope of the GMATvipulgoyal 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
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
Brent Hanneson - Creator of GMATPrepNow.com
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tanvis1120
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Brent@GMATPrepNow
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See the above posts.tanvis1120 wrote:What is insert stick and separator technique?
Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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GMATinsight
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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
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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