Find the Exact Number of Integer Solutions to the Equation
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Find the number of integer solutions to |a| + |b| + |c| = 10, where no [#permalink] 
08 Oct 2019, 16:08
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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
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Re: Find the number of integer solutions to |a| + |b| + |c| = 10, where no [#permalink] 09 Oct 2019, 09:48
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
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Re: Find the number of integer solutions to |a| + |b| + |c| = 10, where no [#permalink] 09 Oct 2019, 12:09
|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
Kinshook wrote:
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
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Re: Find the number of integer solutions to |a| + |b| + |c| = 10, where no [#permalink] 13 Oct 2019, 00:33
nick1816 wrote:
|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
Kinshook wrote:
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
Hello, could you please explain why you're doing 9C2 instead of 9C3?
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Re: Find the number of integer solutions to |a| + |b| + |c| = 10, where no [#permalink] 13 Oct 2019, 02:35
ShreyasJavahar wrote:
nick1816 wrote:
|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
Kinshook wrote:
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
Hello, could you please explain why you're doing 9C2 instead of 9C3?
Because the formula is n-1cr-1
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Re: Find the number of integer solutions to |a| + |b| + |c| = 10, where no [#permalink] 07 Jan 2021, 21:23
nick1816 wrote:
|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
Kinshook wrote:
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
Whats the reasoning behind 9c2, can you please explain?
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Re: Find the number of integer solutions to |a| + |b| + |c| = 10, where no [#permalink] 16 Jan 2021, 16:56
Whats the reasoning behind 9c2, can you please explain?
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Re: Find the number of integer solutions to |a| + |b| + |c| = 10, where no [#permalink] 21 Jan 2021, 13:32
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
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Re: Find the number of integer solutions to |a| + |b| + |c| = 10, where no [#permalink] 11 Jun 2021, 14:50
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
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Re: Find the number of integer solutions to |a| + |b| + |c| = 10, where no [#permalink] 12 Jun 2021, 07:59
[quote="Kinshook"]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
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Find the number of integer solutions to |a| + |b| + |c| = 10, where no [#permalink] 12 Jun 2021, 08:16
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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Find the number of integer solutions to |a| + |b| + |c| = 10, where no [#permalink] 21 Oct 2021, 00:13
(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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Find the number of integer solutions to |a| + |b| + |c| = 10, where no [#permalink]
21 Oct 2021, 00:13
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Find the Exact Number of Integer Solutions to the Equation
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