If (x_1 + x_2 + x_3)(y_1 + y_2) = 77, what is the number of positive

Question

If  (x_1 + x_2 + x_3)(y_1 + y_2) = 77 , what is the number of positive integer solutions of the equation?

(A) 150 
(B) 270
(C) 420 
(D) 1024 
(E) 1048

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Number of positive integral solution
satisfying the equation (x1 + x2 + x3) (y1 + y2) =
77, is
(A) 150 (B) 270
(C) 420 (D) 1024

chetan2u · Accepted Answer

(x_1 + x_2 + x_3)(y_1 + y_2) = 77=7*11 
We are looking for positive integral solutions, meaning none of them will be 0 or negative.

So two cases can be

1)  (x_1 + x_2 + x_3)=7 , that is we have to find ways to distribute 7 things in three parts => (7-1)C(3-1)=6C2=15
  (y_1 + y_2) = 11 , => (11-1)C(2-1)=10C1=10
Total ways = 15*10=150

2)  (x_1 + x_2 + x_3)=11 , that is we have to find ways to distribute 11 things in three parts => (11-1)C(3-1)=10C2=45
  (y_1 + y_2) = 7 , => (7-1)C(2-1)=6C1=6
Total ways = 45*6=270

Total ways = 150+270=420.

You can also find solutions separately for each equation but it will be time consuming 
 (y_1 + y_2) = 11 
Now possibilities are (10,1); (9,2) and so on
Also  (x_1 + x_2+x_3) = 11 
Here possibilities will be 9-1-1 and ways to fill up 3!/2! and so on.

Guntabulla · Answer

Can you explain the logic of making it 6C2 instead of 7C3 which is the typical way of distributing 7 items 3 ways

Oppenheimer1945 · Answer

77=7*11 (both primes)

ab=7*11, (a,b)=(7,11) or (11,7)
=> x1+x2+x3=7 and y1+y2=11 
7-1C3-1 x 11-1C2-1=6C2*10C1=150

or 
=> x1+x2+x3=11 and y1+y2=7 
11-1C3-1 x 7-1C2-1=10C2*6C1=270

Ans=150+270=420

FrozenDragon · Answer

what about case when (x1+x2+x3) is 77 and (y1+y2) is 1

Guntabulla · Answer

y1+y2 cannot be 1 since no two positive integers add together to make 1

Regor60 · Answer

Allocating a number of identical objects, in this case 7, to distinct buckets, in this case 3, such that the buckets add up to the number of objects is not what you describe above.

The approach using the above numbers is:

7+(3-1)C(3-1) = 9C2 or

9!/2!7!

However, this permits scenarios where one or two buckets contain 0 items.

Since the problem stipulates that each bucket must be positive, an adjustment has to be made.

The approach here is then to initially allocate 1 item to each bucket, leaving 4 to be allocated.

Then the approach above is followed:

4+(3-1)C(3-1) = 6C2 = 6!/2!4! = 15

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If (x_1 + x_2 + x_3)(y_1 + y_2) = 77, what is the number of positive

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If (x_1 + x_2 + x_3)(y_1 + y_2) = 77, what is the number of positive

Prep Toolkit

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