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P&C..no. of integral solution

Expert replies
by parveen110 » Sat Jan 18, 2014 8:27 am
Hi,

The number of integral solutions for the equation a+b+c+d=12, Where (a,b,c,d)>=-1 is:

a. 19C3(OA)
b. 18C4
c. 20C4
d. none of these

Please elaborate the approach..Thanks!!
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Source: — Problem Solving |

by Uva@90 » Sat Jan 18, 2014 8:39 am
parveen110 wrote:Hi,

The number of integral solutions for the equation a+b+c+d=12, Where (a,b,c,d)>=-1 is:

a. 19C3(OA)
b. 18C4
c. 20C4
d. none of these

Please elaborate the approach..Thanks!!
Hi Parveen110,
Is this Gmat question ?

Regards,
Uva.
Known is a drop Unknown is an Ocean
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by Brent@GMATPrepNow » Sat Jan 18, 2014 10:03 am
parveen110 wrote:Hi,

The number of integral solutions for the equation a+b+c+d=12, Where (a,b,c,d)>=-1 is:

a. 19C3(OA)
b. 18C4
c. 20C4
d. none of these

Please elaborate the approach..Thanks!!
Hey Parveen,

What's the source of this question?
The questions that you have posted so far are either out of scope or flirting with being out of scope for the GMAT.

Even if the above question were such that a, b, c and d were greater than or equal to zero (in which case the answer would be 15C3), I believe the question would be out of scope. However, once you allow for the variables to equal -1, the question is definitely out of range for the GMAT.

Also, I doubt that the GMAT would assume that everyone is familiar with the nCk notation found in your answer choices. In fact, the Official Guide uses a different (and also acceptable) notation that looks like this:
Image

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
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by GMATGuruNY » Sat Jan 18, 2014 1:28 pm
I agree with Brent that the posted problem is beyond the scope of the GMAT.
That said, here is a similar problem, along with my solution:
How many of the four-digit numbers with non-zero digits have the sum of their digits as 12?
165
330
132
440
170
A sum of 12 units is to be DISTRIBUTED among 4 digits.
None of the digits can be greater than 9.
This is the equivalent of the following problem:

How many ways can 12 identical chocolates be distributed among 4 children A, B, C and D if no child can receive more than 9 chocolates?

To guarantee that no child receives more than 9 chocolates, first distribute 1 chocolate to each child.
Now count the number of ways to distribute the REMAINING 8 CHOCOLATES to the 4 children.

The following is called the SEPARATOR method.

8 identical chocolates are to be separated into -- at most -- 4 groupings.
Thus, we need 8 chocolates and 3 separators:
OO|OO|OO|OO

Each arrangement of the elements above represents one way to distribute the 8 chocolates among 4 children A, B, C and D:
OO|OO|OO|OO = A gets 2 chocolates, B gets 2 chocolates, C gets 2 chocolates, D gets 2 chocolates.
OOOOOO|||OO = A gets 6 chocolates, D gets 2 chocolates.
OOOOOOOO||| = A gets all 8 chocolates.
And so on.

To count all of the possible distributions, we simply need to count the number of ways to arrange the 11 elements above (the 8 identical chocolates and the 3 identical separators).
The number of ways to arrange 11 elements = 11!.
But when an arrangement includes identical elements, we must divide by the number of ways to arrange the identical elements.
The reason:
When the identical elements swap positions, the arrangement doesn't change, reducing the total number of unique arrangements.
Thus, we must divide by 8! (the number of ways to arrange the 8 identical chocolates) and by 3! (the number of ways to arrange the 3 identical separators):
11!/(8!3!) = 165.

The correct answer is A.

For two similar problems, check here:

https://www.beatthegmat.com/algebra-t215423.html#622071
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