If x, y, and z are positive integers, and x + y + z = 9. How many combinations of x, y, and z are possible?
OA 28
combination
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If you like see the approach, please join my upcoming combinatorics session.
You will get to work on a lot more of these questions in my session and in the additional homework that I will be giving out.
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I won't require you to attend a seminar before providing the solution. Who is that guy/girl and who let him/her into our fine community? Heh.shibal wrote:If x, y, and z are positive integers, and x + y + z = 9. How many combinations of x, y, and z are possible?
OA 28
Let's think about the different ways we can sum 3 positive integers to 9:
1/1/7
1/2/6
1/3/5
1/4/4
2/2/5
2/3/4
3/3/3
Now let's think about how many variations of x, y and z those combinations give us:
1/1/7, 1/4/4 and 2/2/5. The single digit in each could be the x, y or z, so each of these gives us 3 possible assignments of numbers to variables. So, that's 9 possibilities so far.
1/2/6, 1/3/5 and 2/3/4. Three distinct numbers can be arranged 3!=6 different ways. So, each of these gives us 6 possible assignments of numbers to variables. That's another 18 possibilities.
3/3/3 can only be assigned 1 way (x=3, y=3, z=3), so that's 1 more possibility.
So, 9 + 18 + 1 = 28 total possibilities.
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I'm not hesitant to give out the solution.Stuart Kovinsky wrote:I won't require you to attend a seminar before providing the solution. Who is that guy/girl and who let him/her into our fine community? Heh.shibal wrote:If x, y, and z are positive integers, and x + y + z = 9. How many combinations of x, y, and z are possible?
OA 28
Let's think about the different ways we can sum 3 positive integers to 9:
1/1/7
1/2/6
1/3/5
1/4/4
2/2/5
2/3/4
3/3/3
Now let's think about how many variations of x, y and z those combinations give us:
1/1/7, 1/4/4 and 2/2/5. The single digit in each could be the x, y or z, so each of these gives us 3 possible assignments of numbers to variables. So, that's 9 possibilities so far.
1/2/6, 1/3/5 and 2/3/4. Three distinct numbers can be arranged 3!=6 different ways. So, each of these gives us 6 possible assignments of numbers to variables. That's another 18 possibilities.
3/3/3 can only be assigned 1 way (x=3, y=3, z=3), so that's 1 more possibility.
So, 9 + 18 + 1 = 28 total possibilities.
You can get the answer in one step. 8C2 = 28.
In order to illustrate the process and setup, it is much easier to do so in a classroom environment. Please understand.
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This is How I did it..
x+y+z=9
x+y=9-z
now since x,y and z are positive integers x+y>=2
so z can take values from 1 to 7
for z=7,x=y=1= 1 solution
for z=6,X+Y=3, x=1,y=2, 2 solutions
do this for a few more and you will see that number ofsolutions are 1,2,3,4 upto 7
total number of solutions=1+..+7=7*8/2=28
But I would like to see if someone has an easier solution
x+y+z=9
x+y=9-z
now since x,y and z are positive integers x+y>=2
so z can take values from 1 to 7
for z=7,x=y=1= 1 solution
for z=6,X+Y=3, x=1,y=2, 2 solutions
do this for a few more and you will see that number ofsolutions are 1,2,3,4 upto 7
total number of solutions=1+..+7=7*8/2=28
But I would like to see if someone has an easier solution
The powers of two are bloody impolite!!
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GMATQuantCoach wrote:I'm not hesitant to give out the solution.Stuart Kovinsky wrote:I won't require you to attend a seminar before providing the solution. Who is that guy/girl and who let him/her into our fine community? Heh.shibal wrote:If x, y, and z are positive integers, and x + y + z = 9. How many combinations of x, y, and z are possible?
OA 28
Let's think about the different ways we can sum 3 positive integers to 9:
1/1/7
1/2/6
1/3/5
1/4/4
2/2/5
2/3/4
3/3/3
Now let's think about how many variations of x, y and z those combinations give us:
1/1/7, 1/4/4 and 2/2/5. The single digit in each could be the x, y or z, so each of these gives us 3 possible assignments of numbers to variables. So, that's 9 possibilities so far.
1/2/6, 1/3/5 and 2/3/4. Three distinct numbers can be arranged 3!=6 different ways. So, each of these gives us 6 possible assignments of numbers to variables. That's another 18 possibilities.
3/3/3 can only be assigned 1 way (x=3, y=3, z=3), so that's 1 more possibility.
So, 9 + 18 + 1 = 28 total possibilities.
You can get the answer in one step. 8C2 = 28.
In order to illustrate the process and setup, it is much easier to do so in a classroom environment. Please understand.
GMACouch:
Anyway u say it, you're conditioning the answer to your seminar, not the objective of this forum, this is not to get students for our courses
Silvia
I think the question can be answered as follows:shibal wrote:If x, y, and z are positive integers, and x + y + z = 9. How many combinations of x, y, and z are possible?
OA 28
Imagine three buckets called x, y, and z. You have to keep at least one fruit in each bucket so that total no of fruits in the buckets is nine. ("at least one" for each of them is positive integers). So better put one fruit to each of buckets. Now you have left with 9-3 = 6 fruits.
Now the question is how these 6 fruits can be distributed?
You line them up ( as shown in figure). Now you have to put two divider in between them to make three groups to find the no. of ways that will attribute to find no. of ways x,y,z combination is possible.
Fruits: F F F F F F
Divider: D D
So a few sequence, randomly taken, may look like:
x D y D z Sequence
F D F D FFFF FDFDFFFF
FF D FF D FF FFDFFDFF
nil D FFF D FFF DFFFDFFF
FFFFFF D Nil D nil FFFFFFDD
and so on.
So that means we have to find no ways two objects can be chosen out of 6(F)+2(D) places i.e. 8C2 = 28
That may help.......
Last edited by real2008 on Thu Jul 23, 2009 9:09 am, edited 1 time in total.
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There are quite a few good ways to do this kind of problem. One way is to imagine nine donuts:
O O O O O O O O O
Now, say we put two partitions ('|') among our donuts:
O O | O O O | O O O O
We've just worked out a way to add three positive integers to get a sum of 9 (just count the donuts in each zone): 2 + 3 + 4 = 9.
For each different partition we can make, we'll get a different set of values for x, y and z that add to 9. There are 8 gaps between pairs of donuts where we could put a partition, so there are 8C2 = 8*7/2 = 28 different ways to choose positive integers for x, y and z so that x+y+z = 9.
O O O O O O O O O
Now, say we put two partitions ('|') among our donuts:
O O | O O O | O O O O
We've just worked out a way to add three positive integers to get a sum of 9 (just count the donuts in each zone): 2 + 3 + 4 = 9.
For each different partition we can make, we'll get a different set of values for x, y and z that add to 9. There are 8 gaps between pairs of donuts where we could put a partition, so there are 8C2 = 8*7/2 = 28 different ways to choose positive integers for x, y and z so that x+y+z = 9.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com
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Good way of thinkingIan Stewart wrote:There are quite a few good ways to do this kind of problem. One way is to imagine nine donuts:
O O O O O O O O O
Now, say we put two partitions ('|') among our donuts:
O O | O O O | O O O O
We've just worked out a way to add three positive integers to get a sum of 9 (just count the donuts in each zone): 2 + 3 + 4 = 9.
For each different partition we can make, we'll get a different set of values for x, y and z that add to 9. There are 8 gaps between pairs of donuts where we could put a partition, so there are 8C2 = 8*7/2 = 28 different ways to choose positive integers for x, y and z so that x+y+z = 9.
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GMATQuantCoach, I've sent you a PM. This is a warning, do not spam our community about your seminar in these threads. You can mention your seminar in the Lounge area, but that's it.
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Ian Stewart, this is by far one of the most beautiful solutions I've seen for combinatoric problems, and I took upper div combo class
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How can I solve the following question using the trick provided in this read?
How many positive integers less than 10,000 are there in which sum of digits equals 5?
(a) 31
(b) 51
(c) 56
(d) 62
(e) 93
How many positive integers less than 10,000 are there in which sum of digits equals 5?
(a) 31
(b) 51
(c) 56
(d) 62
(e) 93
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We can use the same trick, but it's a bit more complicated since we can also use 0 in this question (and in the previous question we knew that x, y and z were all positive).shanrizvi wrote:How can I solve the following question using the trick provided in this read?
How many positive integers less than 10,000 are there in which sum of digits equals 5?
(a) 31
(b) 51
(c) 56
(d) 62
(e) 93
Here, we have 4 digits (positive integer less than 10000 means that 9999 is the biggest and we can pretend that "5" is "0005") that must sum to 5.
Since we have 4 digits, we'll have 3 partitions. We're summing to 5, so we have 5 "donuts".
O O O O O
Since we can use 0, we can have multiple partitions in the same spot. For example, we could have:
|||OOOOO (which translates to 0005)
we could have:
||O|OOOO (which translates to 0014, or 14).
So, we view this as a permutation question: we have 8 total objects, 3 of which are identical to each other (the partitions) and 5 of which are identical to each other (the donuts). Using the permutation formula for which some objects are identical:
Total permutations = n!/r!s!
Total permutations = 8!/3!5! = 8*7*6/3*2*1 = 8*7 = 56... choose (C).
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Please, what does c mean in 8C2? does it matter where you put the partition in any problem ?Ian Stewart wrote:There are quite a few good ways to do this kind of problem. One way is to imagine nine donuts:
O O O O O O O O O
Now, say we put two partitions ('|') among our donuts:
O O | O O O | O O O O
We've just worked out a way to add three positive integers to get a sum of 9 (just count the donuts in each zone): 2 + 3 + 4 = 9.
For each different partition we can make, we'll get a different set of values for x, y and z that add to 9. There are 8 gaps between pairs of donuts where we could put a partition, so there are 8C2 = 8*7/2 = 28 different ways to choose positive integers for x, y and z so that x+y+z = 9.
Thank you