The number of positive integral solutions of the equation x+y+z+t+e =

Question

The number of positive integral solutions of the equation x+y+z+t+e = 10 is
A.60
B.130
C.126
D.90
E.91

CrackverbalGMAT · Accepted Answer

For any equation of the form a + b + c +... till r variables = n

Then the total number of solutions =  ^{n-1}C_{r-1}

Here n = 10 and r = 5

Therefore total solutions =  ^{10-1} C _ {5 - 1}  =  ^9C_4 = \frac{9!}{(9-4)!*4!} = \frac{9!}{(5)!*4!} = \frac{9 * 8 * 7 * 6 * 5!}{5! * 4 * 3 * 2 * 1} = \frac{3024}{24} = 126

Option C

Arun Kumar

IanStewart · Answer

This is known as a "partition problem" in combinatorics. I've never seen a partition problem in any official GMAT material, so it's very unlikely anyone will need to understand how to solve this kind of problem. But if we take 10 dots:

• • • • • • • • • •

then if we insert four 'partitions' between different pairs of dots, for example like this:

• | • • • | • •| • | • • •

then we'll be finding a way to write "10" as a sum of five numbers. So in the example above, counting the dots in each zone from left to right, we're showing that 10 = 1 + 3 + 2 + 1 + 3. Any of the ways we can insert four partitions will give us a different way to arrive at "10" as a sum of five numbers, and since there are 9 spaces between dots, and we need to insert 4 partitions, the answer is 9C4 (since order does not matter - switching one partition symbol with another doesn't change how we're dividing the dots). So the answer is (9)(8)(7)(6)/4! = (9)(7)(2) = 126.

rajatchopra1994 · Answer

Explanation:

x+y+z+t+e = 10
then (x-1) + (y-1) + (z-1) + (t-1) + (e-1) = 5.
Set a=(x-1), b=(y-1), c=(z-1), d=(t-1) ,f=(e-1). 
a+b+c+d+f=5
(n+r-1)C(r-1)
n=5 , r=5
9C4 = 126

IMO-C

Aulegue · Answer

The word problem version of the above:

In how many ways can we distribute 10 candies (=identical things) amongst 5 children so that each child receives at least one (since we consider only positive solutions) candy?

Firstly, give one to each child --> 5 candies are left
Secondly, distribute 5 left candies among 5 children; any child may get any number of candies, incl. 0 
(5+ )C( )=9!/(5!*4!)=9*8*7*6/4*3*2=126

Why  ? Because we need only 4 dots to get 5s line segments (have a look at Ian's post). And then a problem becomes exactly like a COFFEE/SUCCESS/HHHTT, etc. type of problem.

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The number of positive integral solutions of the equation x+y+z+t+e =

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