# Thread: find the number of different order triples

1. ## find the number of different order triples

Find the number of different order triples (a,b,c) s.t a+b+c=12 if

a) a,b,c must be nonegetive integers

b) a,b,c must be positive integers

is this permatutaion or combination? Suppose there is 12 choices for a , then b will be depend on a, and c will depend on both a and b

2. ## Re: find the number of different order triples

Hello, wopashui!

Believe it or not this is a Combination problem.
But you'll have see the explanation.

Find the number of different ordered triples $(a,b,c)$
such that $a+b+c\,=\,12$ if:

. . a) $a,b,c$ must be nonnegetive integers.

Place 12 objects in a row, inserting a space before, after and between them.

. . $\_\;*\;\_\;*\;\_\;*\;\_\;*\;\_\;*\;\_\;*\;\_\;* \;\_\;*\;\_\;*\;\_\;*\;\_\;*\;\_\;*\;\_$

Place two "dividers" in any of the 13 spaces.

So that: . $*\;*\;*\;*\;*\,|\,*\;*\;*\;*\,|\,*\;*\;\:*$ .represents: $5 + 4 + 3$

. . and: . $*\;*\;*\;*\,||\,*\;*\;*\;*\;*\;*\;*\;\,*$ .represents: $4 + 0 + 8$

If the dividers are placed in two different spaces, there are: . $_{13}C_2 = 78$ ways.

If the dividers are placed in the same space, there are: $13$ ways.

Hence, there are: . $78 + 13 \,=\,91$ ways to place the dividers.

Therefore, there are $91$ triples of nonnegative integers whose sum is 12.

b) $a,b,c$ must be positive integers.

Place 12 objects in a row, inserting a space between them.

. . $*\;\_\;*\;\_\;*\;\_\;*\;\_\;*\;\_\;*\;\_\;* \;\_\;*\;\_\;*\;\_\;*\;\_\;*\;\_\;*$

Choose 2 of the 11 spaces and insert "dividers".

So that: . $*\;*\;*\;*\;*\,|\,*\;*\;*\;*\,|\,*\;*\;\:*$ .represents: $5 + 4 + 3$

. . and: . $*\;*\;*\;*\,|\,*\;*\;*\;*\;*\;*\;*\,|\,*$ .represents: $4 + 7 + 1$

The dividers are placed in two different spaces.
. . There are: . $_{11}C_2 = 55$ ways.

Therefore, there are $55$ ordered triples of positive integers whose sum is 12.

3. ## Re: find the number of different order triples

thanks alot, but i dun understand why is the second part only has 11C2, not 13C2?

4. ## Re: find the number of different order triples

anyone can explain?

5. ## Re: find the number of different order triples

Originally Posted by wopashui
Find the number of different order triples (a,b,c) s.t a+b+c=12 if
b) a,b,c must be positive integers
Originally Posted by wopashui
thanks alot, but i dun understand why is the second part only has 11C2, not 13C2?
Think of the 12 as being twelve 1's. Lets make the variables positive by putting a 1 is each to begin with. Now that leaves us with nine 1's to distribute: $\binom{9+3-1}{9}$.

6. ## Re: find the number of different order triples

thx,, so the first part would be 12+3-1 C 14, which is 91

7. ## Re: find the number of different order triples

Originally Posted by wopashui
thx,, so the first part would be 12+3-1 C 14, which is 91
@wopashui,
You have absolutely no idea about any of this.
You are either playing us for fools or you just don't get it.
I am done with you in any case.
Good luck.