Find the number of integer solutions to |a| + |b| + |c| = 10, where no

Question

Find the number of integer solutions to |a| + |b| + |c| = 10, where none of a, b or c is 0.

A. 36

B. 72

C. 144

D. 288

E. 576

MahmoudFawzy · Accepted Answer

First, lets assume that a,b,c are all +ve to simplify the question. 
The three numbers can have the following combinations:
118 --> with 3 rearrangements ( because 1 is repeated, the possibilities =  \frac{3!}{2!} = 3 )
127 --> with 6 rearrangements (because all numbers are different, the possibilities =  3! = 6 )
136 --> with 6 rearrangements
145 --> with 6 rearrangements
226 --> with 3 rearrangements
235 --> with 6 rearrangements
244 --> with 3 rearrangements
334 --> with 3 rearrangements

Total = 36
( the above part is similar to another interesting counting question mentioned here : ( https://gmatclub.com/forum/in-how-many-ways-10-identical-chocolates-be-distributed-among-3-child-307384.html#p2375066 )

However, each of a,b,c can be either +ve or -ve, 
so the combined possibilities of the signs = 2*2*2 = 8
for details, (a,b,c) can be:
(+,+,+)
(+,+,-)
(+,-,+)
(-,+,+)
(+,-,-)
(-,+,-)
(-,-,+)
(-,-,-)
8 possibilities

so the total number of possibilities = 36*8 = 288
 D

IanStewart · Answer

This type of problem, where essentially we're trying to find how many positive integer solutions there are to an equation like "a + b + c = 10" (or any other equation where positive integers sum to a number) is known as a "partition problem" in combinatorics. I've never once seen a partition problem in official GMAT materials, so it's not likely this will be important to understand. If we draw ten dots:

• • • • • • • • • •

and then we insert two partitions in random places, between different pairs of dots, e.g. like this:

• • • | • • | • • • • •

then, counting dots on each side of a partition, we create a solution to the equation a + b + c = 10. So the diagram above illustrates the solution 3 + 2 + 5 = 10. The number of different positive integer solutions to "a + b + c = 10" will equal the number of ways we can insert two partitions in the middle of 10 dots. And we have 9 spaces between dots, so 9 choices for where to put the first partition. Since our partitions can't go in the same place (a, b and c cannot be zero, so we must have at least one dot between two partitions), we only have 8 choices for where to put the second partition. But if we flip the positions of our two partition markers, our diagram stays unchanged, so the order of the two partitions doesn't matter, so we must divide by 2!. So we have (9)(8)/2! = 36 ways to place the partitions, and there are thus 36 distinct positive integer solutions to the equation a + b + c = 10.

Since we have two choices for the sign of a, two choices for the sign of b, and two choices for the sign of c if we want nonzero integer solutions to |a| + |b| + |c| = 10, we have (2)(2)(2)(36) = 288 solutions in total.

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nick1816 · Answer

|a|+|b|+|c|=10

integral solutions possible for (|a|, |b|, |c|)= 9C2=36

Now a, b and c can can be positive or negative; Hence total possible solutions for (a,b,c)= 2*2*2*36= 288

ShreyasJavahar · Answer

Hello, could you please explain why you're doing 9C2 instead of 9C3?

Peddi · Answer

Because the formula is n-1cr-1

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RohitSaluja · Answer

Whats the reasoning behind 9c2, can you please explain?

dreamaboutit · Answer

Whats the reasoning behind 9c2, can you please explain?

Iq · Answer

Assume 3 dies first, that add up to 10. Each has numbers from 1 - 6, and the no. of ways to get a sum of 10 on 3 dies = 27 ways

Each no. can be +ve or -ve, so each number can be selected in 2 ways.

3 No.s, each + or -ve adding to 10 = 27 * 2 * 2 * 2 = 216 ways

There will be ONLY two additional possibilities: i.e. (7,2,1) and (8,1,1)

So including this, will increase the total from 216 but definitely not up to 576.

D

PS: Exact numbers can be calculated, but not needed at this point: 
(7,2,1) and (8,1,1) can be re-arranged in : 3! + (3!)/2 ways = 9
each can be positive or negative, so total as above: 9 * 8 = 72
Total solutions = 216 + 72 =  288

Bambi2021 · Answer

Mississippi sort of reasoning here:

(x)x x x x|x x x|x(x)

a, b, c cant be zero. This is indicated by the parentheses (=the bars cant be at the outermost positions). If I arrange these I get 10!/(2!*8!) = 45. In this case I also counted the cases where the bars are adjacent, which is not allowed. As they can be adjacent in 9 ways, I subtract these from 45 to arrive at 36.

For every solution, however, any of the variables can be switched between positive or negative. I account for this by multiplying 36 by 2^3.

36*8 = 288

Kinshook · Answer

Find the number of integer solutions to |a| + |b| + |c| = 10, where none of a, b or c is 0.

Let us first find the solutions for 
a + b + c = 10

(a, b, c) = {(1,1,8),(1,8,1),(8,1,1),(1,2,7),(1,7,2),(2,1,7),(7,1,2),(7,2,1),(2,7,1),(1,3,6),(3,1,6),(1,6,3),(3,6,1),(6,1,3),(6,3,1),(1,4,5),(1,5,4),(4,1,5),(4,5,1),(5,1,4),(5,4,1),(2,2,6),(2,6,2),(6,2,2),(2,3,5),(2,5,3),(3,2,5),(3,5,2),(5,2,3),(5,3,2),(2,4,4),(2,4,2),(4,2,2)(3,3,4),(3,4,3),(4,3,3)} : 36 solutions

We can take negative values for each of a, b, & c as well
Therefore, number of integer solutions = 2*2*2*33 = 8*36 = 288 solutions

IMO D

Fdambro294 · Answer

(1st)

Consider the case where each of A, B, and C must be positive integer solutions

A + B + C = 10

Since none of the integer values can equal zero——>Thus:

A = (a + 1)

B = (b + 1)

C = (c + 1)

(a + 1) + (b + 1) + (c + 1) = 10

a + b + c = 7

To find how many positive integer values satisfy this linear equation, we can use the “stars and bars” method for distributing identical items (in this case, it would be identical values of 1) into distinct groups (in this case, the variables of: a, b , c)

* | * * * | * * *

Each * stands for one value of “1”

And each partition “I” separates the “1’s” into each distinct variable.

For instance the above distribution would be

a = 1

b = 3 (because there are 3 stars in the middle group)

c = 3 (because there are 3 stars on the end group

This would make

A = 2

B = 4

C = 4 ————> where A + B + C = 10

The different arrangements of these identical elements will give us all the different distributions that are possible for positive integers only

(7 + 2)! 
______ =
(2!) (7!)

(9 *8 * 7!)
__________ =
(2!) * (7!)

(9 * 8) / 2 = 36 Ordered Solutions in which (A , B, C) are POSITIVE Integers

(2nd) because there is an absolute value Modulus around each variable, for any one of these 36 ordered solutions, the numbers themselves can vary between (+)pos. and (-)neg.

We can have

Case 1: (+) (+) (+) ————> 1 possibility of 36 arrangements

Case 2: (-) (-) (-) —————> 1 possibility of the 36 arrangements

Case 3: (+) (+) (-) ———> in which 2 variables are positive and 1 variable is negative. For each of the 36 ordered solutions, we can vary the signs in:

(3!) / (2!) = 3 ways

Case 4: (-) (-) (+) ———-> same logic as case 3, except now we have 2 negative and 1 positive variable

3 ways

Total ways = 1 + 1 + 3 + 3 = 8 ways

So if we were to list out the 36 ordered solutions, we could vary the (+) and (-) signs among the 36 ordered solutions in:

(36) (8) = 288 ways

Answer

288

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samarpan.g28 · Answer

Let's make cases for a, b, and c.
a b c
1 1 8 - 3!/2!=3 ways
1 2 7 - 3!=6 ways
1 3 6 - 3!=6 ways
1 4 5 - 3!=6 ways
2 2 6 - 3!/2!=3 ways
2 3 5 - 3!=6 ways
2 4 4 - 3!/2!=3 ways
3 3 4 - 3!/2!=3 ways
Total 36 ways.
Since a, b, and c take mod values, they can all be positive, or one negative, or two negatives, or all three negative i.e. 3C0+3C1+3C2+3C3=8
So, 36*8=288(D).

Oppenheimer1945 · Answer

No of non-int sol of x+y+z=10 is 9C2,
for each variable, we have 2 possibilities(+ or -), hence 2^3 9C2= 288

StillMotor · Answer

Hello, Please explain why we use 2!

egmat · Answer

Hi StillMotor,

The  2!  you're seeing shows up in the case-counting solutions (like MahmoudFawzy's and samarpan's lists), on rows such as  1, 1, 8  or  2, 2, 6 , where the count is written as  3!/2! . Let me clear up exactly why that division happens.

The idea: 3! overcounts when a number repeats

When all three numbers are  different  - say 1, 2, 7 - the number of ways to arrange them across positions (a, b, c) is  3! = 6 . Every rearrangement is genuinely a new, distinct triple.

But when two of the numbers are  the same  - like  1, 1, 8  - 3! counts arrangements that are actually identical. Swapping the two 1's changes nothing, yet 3! treats them as separate. So we divide by  2!  (the number of ways to arrange the two identical 1's among themselves) to erase that double-count:

-  3! / 2! = 6 / 2 = 3  actual distinct arrangements.

That's all the  2!  is doing: it removes the overcount created by the repeated value.

See it by hand

List every arrangement of  1, 1, 8  across positions (a, b, c):

- (1, 1, 8)
- (1, 8, 1)
- (8, 1, 1)

That's exactly  3  - not 6. If you tried to write out 6, you'd be forced to repeat one of these, because the two 1's are indistinguishable. That repetition is the  2!  you're dividing away.

Compare with  1, 2, 7 , which really does give all  6  distinct arrangements - no repeat, so no division needed.

Rule of thumb:  start with 3!, then divide by k! for each group of k identical numbers. That keeps every triple counted exactly once.

Answer: D

BraveVault · Answer

What a beautiful way to explain! Amazing!

Find the number of integer solutions to |a| + |b| + |c| = 10, where no

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