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inserting stick or seperator rule

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inserting stick or seperator rule

by sana.noor » Tue Jul 30, 2013 12:03 pm
If x, y, and z are positive integers, and x + y + z = 9. How many combinations of x, y, and z are possible?

OA is 28

i am using "inserting stick rule. suppose x=3 y= 2 z= 4
324 = ***|**|**** = 9+2!/9! 2! = 55
where i am wrong?
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by Matt@VeritasPrep » Tue Jul 30, 2013 1:12 pm
You're using the right rule -- one that I've never seen tested on the GMAT, BTW, even though it's intuitive and really cool -- but your numbers are a bit off.

Think of your stars this way:

*_*_*_*_*_*_*_*_*

Since you can't have 0 as any of your integers, you have to place your sticks IN BETWEEN your stars. So there are only 8 places your sticks can go (shown above). You have 2 sticks, so (8 choose 2) = 28.
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by GMATGuruNY » Tue Jul 30, 2013 3:47 pm
sana.noor wrote:If x, y, and z are positive integers, and x + y + z = 9. How many combinations of x, y, and z are possible?
A sum of 9 is to be DISTRIBUTED among integers x, y and z.
Since x, y and z must be POSITIVE, each must be greater than or equal to 1.
The problem above is equivalent to the following:
How many ways can 9 identical chocolates to be distributed among 3 people A, B and C, if each person must receive AT LEAST 1 chocolate?
To ensure that each person receives at least 1 chocolate, first give 1 chocolate to each person.
Now we need to count the number of ways to distribute the REMAINING 6 CHOCOLATES among the 3 people.

The following is called the SEPARATOR method.

6 identical chocolates are to be separated into -- at most -- 3 groupings.
Thus, we need 6 chocolates and 2 separators:
OO|OO|OO

Each arrangement of the elements above represents one way to distribute the 6 chocolates among the 3 people A, B and C:
OO|OO|OO = A gets 2 chocolates, B gets 2 chocolates, C gets 2 chocolates.
OO||OOOO = A gets 2 chocolates, B gets 0 chocolates, C gets 4 chocolates.
OOOOOO|| = A gets all 6 chocolates.
And so on.

To count all of the possible distributions, we simply need to count the number of ways to arrange the 8 elements above (the 6 identical chocolates and the 2 identical separators).
The number of ways to arrange 8 elements = 8!.
But when an arrangement includes identical elements, we must divide by the number of ways to arrange the identical elements.
The reason:
When the identical elements swap positions, the arrangement doesn't change, reducing the total number of unique arrangements.
Here, we must divide by the number of ways to arrange the 6 identical chocolates (6!) and the number of ways to arrange the 2 identical separators (2!):
8!/(6!2!) = 28.
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by [email protected] » Tue Jul 30, 2013 4:36 pm
Hi Sana.noor,

Mitch & Matt have both properly explained the high-concept idea behind this question, so I won't rehash that idea. Instead, I'll offer an alternative. On of the toughest things that you have to face on Test Day is the Countdown Clock. You have just 75 minutes to get the Quant section done, and while 75 minutes is enough time to do what you need to do, if you're staring at the screen (and doing nothing), then you're in trouble.

Certain high-concept questions on the GMAT can be beaten rather easily, by children, who have no knowledge of advanced math. Essentially, the children just "brute force" the question. You can do what I'm about to show you and have the solution relatively quickly.

First, how many ways are there for x, y and z to be positive integers and add up to 9?

117
126
135
144
225
234
333

So, there are 7 sets of numbers.

Now, how many ways are there to match each variable (x, y and z) to a number in each set?

When the 3 numbers are different, such as 234, you'd have:
234
243
324
342
423
432
6 ways

When the 3 numbers include a duplicate, such as 225, you'd have:
225
252
522
3 ways

When the 3 numbers are the same, such as 333, you'd have
333
1 way

With this info and the master list of possibilities, you'd have:
117 3 ways
126 6 ways
135 6 ways
144 3 ways
225 3 ways
234 6 ways
333 1 way

Total: 28 ways

Now, I absolutely love knowing all of the shortcuts, but if you're "stuck" or "unsure" of how a concept works, then you have to be ready to shift tactics. When a question asks for the number of possibilities and the options are obvious limited, sometimes you just have to map out all the options.

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by nikhilgmat31 » Wed Oct 07, 2015 12:20 am
We can count it in simple way as
144
234
126
333
522
711
531

144 can arranged in 3!/2! = 3 ways
234 can arranged in 3! = 6 ways
126 can arranged in 3! = 6 ways
333 can arranged in 1! = 1 way
522 can arranged in 3!/2! = 3 ways
711 can arranged in 3!/2! = 3 ways
531 can arranged in 3! = 6 ways

Total number of ways = 28 ways.
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