Counting and Combinatorics

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July 7th 2009, 03:56 PM
billym

Counting and Combinatorics

Q: How many solutions are there to:

x1 + x2 + x3 + x4 +x5 = 50 for integers xi >= 1

What I've done:

Rewrite as: (x1 - 2) + (x2 - 2) + (x3 - 2) + (x4 - 2) + (x5 - 2) = 40

My answer: The number of solutions is C(40 + 5 - 1 , 40 ) = 135751

The answer has been given as 130905. Any idea where I have gone wrong?
July 7th 2009, 05:09 PM
Soroban

Hello, billym!

I don't agree with their answer . . .

Quote:

How many solutions are there to:

. . $x_1 + x_2 + x_3 + x_4 +x_5 \:=\: 50$ . for integers $x_i \geq 1$

The answer has been given as 130,905.

Consider a 50-inch dowel ... marked in one-inch intervals.
. . $\square\square\square\square\square\square\square\ square\hdots\square\square$

There are 49 marks along the dowel.
We will select 4 of them to make our cuts, dividing the dowel into 5 parts.

There are: . $_{49}C_4 \:=\:{49\choose4} \:=\:\frac{49!}{4!\,45!} \;=\; 211,\!876$ choices.

Therefore, there are 211,876 solutions.
July 7th 2009, 05:42 PM
Plato

If each answer is a at least one, then we are dealing with $xL1+xL2+xL3+xL4+xL5=45$ having "put" a one into each variable.
So the solution is $\binom{45+5-1}{45}$ .

Here is the thought experience.
Think of putting a 1 into each of the five variables.
That leaves 45 one’s to place into five different cells: $\binom{45+5-1}{45}$ .
July 7th 2009, 06:06 PM
Random Variable

If you expand $(x+x^{2}+x^{3}+...)^{5}$ , the coefficient of the $x^{50}$ term is 211876
July 7th 2009, 06:07 PM
billym

Hmm...

That was the first, first answer I had:

Rewrite as: (x1 - 1) + (x2 - 1) + (x3 - 1) + (x4 - 1) + (x5 - 1) = 45

Thus, C(45 + 5 - 1 , 40 ) = 211876

Guess the answer sheet is wrong. I hate that.
July 7th 2009, 06:12 PM
Random Variable

Your answer is correct if $x_{i} \ge 2$