# Thread: Generalized Permutations and Combinations

1. ## Generalized Permutations and Combinations

Hello guys,

I have been stuck on this question for soo long ... I was wondering if anyone could help me out here?

The question is:

x1 + x2 + x3 = 14

x1, x2, x3 are non-negative integers.

How many solutions can John find for x1, x2 and x3 if x1 > 3?

Then I have to present the answer in either P(n,r) or C(n,r).

My problem lies in that fact that x1 has (4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14) whereas x2 and x3 has (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14), therefore all the formulas I know can't seem to work it out.

Ordinarily if the range of numbers were the same for all 3 numbers I would have worked it out like this:

C(14+3-1, 3-1) = C(16, 2)

Any help would be greatly appreciated.
Thanks.

Edit: I finally worked it out thanks to the almighty Moo !!

What I had to do was very simple !

Since there are 4 integers missing from the front of the x1 set, its impossible to have the last 4 integers in the x2 and x3 sets .. therefore it evens out the sets .. and then its a simple tallying up however many valid numbers are left and doing filling in the below:

C(10 + 3 - 1, 3 - 1)

C = Combination as I think this is an unordered selection with repetitions allowed.
10 = n (number of valid numbers)
3 - 1 = r (for x1, x2, x3)

I'm probably not that great at explaining it .. but I worked it out !! haha !

Hello guys,

I have been stuck on this question for soo long ... I was wondering if anyone could help me out here?

The question is:

x1 + x2 + x3 = 14

x1, x2, x3 are non-negative integers.

How many solutions can John find for x1, x2 and x3 if x1 > 3?

Then I have to present the answer in either P(n,r) or C(n,r).

My problem lies in that fact that x1 has (4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14) whereas x2 and x3 has (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14), therefore all the formulas I know can't seem to work it out.

Ordinarily if the range of numbers were the same for all 3 numbers I would have worked it out like this:

C(14+3-1, 3-1) = C(16, 2)

Any help would be greatly appreciated.
Thanks.

Edit: I finally worked it out thanks to the almighty Moo !!

What I had to do was very simple !

Since there are 4 integers missing from the front of the x1 set, its impossible to have the last 4 integers in the x2 and x3 sets .. therefore it evens out the sets .. and then its a simple tallying up however many valid numbers are left and doing filling in the below:

C(10 + 3 - 1, 3 - 1)

C = Combination as I think this is an unordered selection with repetitions allowed.
10 = n (number of valid numbers)
3 - 1 = r (for x1, x2, x3)

I'm probably not that great at explaining it .. but I worked it out !! haha !
Correct.

Note that the solution can also be given as the sum of the arithmetic series 1 + 2 + ... + (14 - 3):

$x_1 = 4 \Rightarrow x_2 + x_3 = 10$ clearly gives 11 solutions: (4, 0, 10), (4, 1, 9), ....... (4, 10, 0).

$x_1 = 5 \Rightarrow x_2 + x_3 = 9$ clearly gives 10 solutions: (5, 0, 9), (5, 1, 8), ....... (5, 9, 0).

etc.

$x_1 = 14 \Rightarrow x_2 + x_3 = 0$ clearly gives 1 solution: (14, 0, 0).

This is unsurprising since $\frac{(14-3)(14-2)}{2} = \frac{(11)(12)}{2} = {12 \choose 2} = {10 + 3 - 1 \choose 3 - 1}$ .....