Thread: comb or perm?

From the set of numbers: {1, 2, 3, 4, 5, 6}, how many different sums can be formed by summing up any two numbers in the set?

I'm thinking this could be a permutation because 1+2 and 2+1 are not two different sums. So, $\displaystyle 6P2$=30.

Is this correct? Or would it be a combination...

Originally Posted by sfspitfire23

From the set of numbers: {1, 2, 3, 4, 5, 6}, how many different sums can be formed by summing up any two numbers in the set?

I'm thinking this could be a permutation because 1+2 and 2+1 are not two different sums. So, $\displaystyle 6P2$=30.

Is this correct? Or would it be a combination...

You counted the number of ways to arrange sets of two digits (permutations), which include 1,2 and 2,1.

Instead, since 1+2=2+1, you should count the number of combinations (or selections).

You should count neither permutations nor combinations, because some distinct combinations give the same sum. For example, 2+4 = 1+5.

So it's not a permutation or combination problem, it's a "think about it and figure out the answer" problem.

Yes,
count combinations if you want the sums of different numbers,
but not if you want different answers!

If you add 2 adjacent numbers in your list,
they will equal the sum of the numbers that "flank" them,

ie 2+3=1+4, 3+4=2+5, 4+5=3+6

if a number has 2 numbers to the left and 2 numbers to the right,
you get (flanking 3)...2+4=5+1, (flanking 4)...3+5=6+2.

Flanking both 3 and 4...2+5=6+1

Originally Posted by sfspitfire23

[B]From the set of numbers: {1, 2, 3, 4, 5, 6}, how many different sums can be formed by summing up any two numbers in the set?
would it be a combination...

Again, as has been pointed out, it is neither as combination nor permutation problem.
If we add two of those numbers the minimum sum is 3 and the maximum is 11.
So how many different sums are there?

Hello, sfspitfire23!

From the set of numbers: {1, 2, 3, 4, 5, 6},
how many different sums can be formed by adding any two numbers in the set?

Assuming numbers can be repeated, we have this addition table:

. . $\displaystyle \begin{array}{c|cccccc}
+ & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline
1 & 2 & 3 & 4 & 5 & 6 & 7 \\
2 & 3 & 4 & 5 & 6 & 7 & 8 \\
3 & 4 & 5 & 6 & 7 & 8 & 9 \\
4 & 5 & 6 & 7 & 8 & 9 & 10 \\
5 & 6 & 7 & 8 & 9 & 10 & 11 \\
6 & 7 & 8 & 9 & 10 & 11 & 12
\end{array}$

There are eleven different sums (2 through 12).



If numbers can not be repeated, we have this table:

. . $\displaystyle \begin{array}{c|cccccc}
+ & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline
1 & - & 3 & 4 & 5 & 6 & 7 \\
2 & 3 & - & 5 & 6 & 7 & 8 \\
3 & 4 & 5 & - & 7 & 8 & 9 \\
4 & 5 & 6 & 7 & - & 9 & 10 \\
5 & 6 & 7 & 8 & 9 & - & 11 \\
6 & 7 & 8 & 9 & 10 & 11 & -
\end{array}$

There are nine different sums (3 through 11).

Originally Posted by Plato

Again, as has been pointed out, it is neither as combination nor permutation problem.
If we add two of those numbers the minimum sum is 3 and the maximum is 11.
So how many different sums are there?

More of the same!!
Depending on the answer being sought,

the question may be solved with combinations if desired.

$\displaystyle \binom{6}{2}-6$