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by GMATQuantCoach » Wed Jul 22, 2009 4:05 pm
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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Re: combination

by Stuart@KaplanGMAT » Wed Jul 22, 2009 7:11 pm
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
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.

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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Re: combination

by GMATQuantCoach » Wed Jul 22, 2009 7:32 pm
Stuart Kovinsky wrote:
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
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.

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.
I'm not hesitant to give out the solution.

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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by tohellandback » Wed Jul 22, 2009 7:34 pm
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
The powers of two are bloody impolite!!
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Re: combination

by ssuarezo » Wed Jul 22, 2009 8:20 pm
GMATQuantCoach wrote:
Stuart Kovinsky wrote:
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
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.

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.
I'm not hesitant to give out the solution.

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
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Re: combination

by real2008 » Thu Jul 23, 2009 1:57 am
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
I think the question can be answered as follows:

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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by Ian Stewart » Thu Jul 23, 2009 3:10 am
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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by way2kailash » Thu Jul 23, 2009 3:28 am
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.
Good way of thinking
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by beatthegmat » Thu Jul 23, 2009 4:05 am
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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by PussInBoots » Fri Jul 24, 2009 10:27 pm
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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by shanrizvi » Tue Aug 25, 2009 2:22 am
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
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by Stuart@KaplanGMAT » Tue Aug 25, 2009 1:23 pm
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
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).

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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by heshamelaziry » Sat Aug 29, 2009 2:35 pm
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.
Please, what does c mean in 8C2? does it matter where you put the partition in any problem ?

Thank you
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