Neat counting problem

Neil30z · New Member Offline Joined: 10 Feb 2009 Posts: 13

In the equation:
x + y + z + w = 17
x, y, z, w, are all positive integers.

How many ordered quadruples (x,y,z,w) are there that satisfy the equation?

modularmarc101

Let 17 dots * * * ... * * represent the 17 in the equation. We can separate them with 3 dividers to create 4 regions that represent the values of the 4 variables. There are 16 spaces available to place the dividers, so there are $\binom{16}{3} = \boxed{560}$ solutions.

**Posted:** Sat Nov 07, 2009 7:27 pm

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Neil30z · New Member Offline Joined: 10 Feb 2009 Posts: 13

You had the right reasoning/intuition about it, but you made a mistake about how to think of the string of 17 dots. Remember, if you have the 3 dividers there, they add to the amount of "dots" in the string.

**Posted:** Sat Nov 07, 2009 7:39 pm

modularmarc101

I don't understand. What do you mean?

**Posted:** Sat Nov 07, 2009 7:42 pm

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Neil30z · New Member Offline Joined: 10 Feb 2009 Posts: 13

Take a look at the 17 dots you have. As you said, there should be 3 dividers. So in your string of 17 dots, add 3 more dots, but make them look different, and they'll denote the dividers. So in total, your string now contains 20 characters. Now look at the repeating characters that you have. This just becomes a factorial problem.

**Posted:** Sat Nov 07, 2009 7:46 pm

modularmarc101

Ok, but the dividers can not be at the ends and they can not be next to each other, so it is $16 * 14 * 12 = 2688$ ?

According to what you're saying it should be $\frac{20!}{17!3!} = 1140$ ?

**Posted:** Sat Nov 07, 2009 7:52 pm

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MathAndKnowledge

Neil30z you are wrong because x, y, z, and w are positive integers. If they were nonnegative integers then your reasoning would hold. modularmarc101's initial answer is correct.

**Posted:** Sat Nov 07, 2009 7:55 pm

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Neil30z · New Member Offline Joined: 10 Feb 2009 Posts: 13

**Posted:** Sat Nov 07, 2009 8:12 pm

Ihatepie

For anyone confused, modular had the right answer the first time, but for the wrong reasons.

**Posted:** Sat Nov 07, 2009 9:52 pm

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Neil30z · New Member Offline Joined: 10 Feb 2009 Posts: 13

You could just do 20 choose 3 and get 1140, because first you're adding the 3 dividers to the string, and then you could just choose the different positions of the dividers in that string.

**Posted:** Sun Nov 08, 2009 8:27 am

Eulers_Apprentice · Hodge Conjecture Offline Joined: 05 Aug 2008 Posts: 51

Ihatepie is right, the answer is $560$ .
why?

**Posted:** Sun Nov 08, 2009 3:38 pm

modularmarc101

**Posted:** Sun Nov 08, 2009 5:10 pm

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AIME15

Read the post above yours Wink

**Posted:** Sun Nov 08, 2009 5:16 pm

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modularmarc101

But in that case, the arraw of dots has 14 spaces in total.. not 16 ??

**Posted:** Sun Nov 08, 2009 5:20 pm

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Ihatepie

**Posted:** Sun Nov 08, 2009 5:43 pm

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randomguy64

**Posted:** Sun Nov 08, 2009 7:01 pm

Ihatepie

No, the dividers aren't distinguishable. The spot to the left of the first divider is x, in between 1st and 2nd is y, etc.

**Posted:** Sun Nov 08, 2009 7:35 pm

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randomguy64

Oh. OK.

I get it now. Thanks!

**Posted:** Mon Nov 09, 2009 3:15 pm

Neil30z · New Member Offline Joined: 10 Feb 2009 Posts: 13

Sorry everybody, but I just realized that the problem said positive, therefore modularmarc's first answer is correct. My answer (1140) is for nonnegative.

**Posted:** Wed Nov 11, 2009 3:00 pm