Math Help Forum: comb or perm?

  1. #1
    Member sfspitfire23's Avatar
    Joined
    Oct 2009
    Posts
    243

    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,
    6P2=30.

    Is this correct? Or would it be a combination...
    Follow Math Help Forum on Facebook and Google+

  2. Welcome to Math Help Forum - Click here to Register

    Welcome to the largest Math Help Forum, a free community dedicated to math help and math discussions.

    We welcome everyone and the community is free to join so register today and become part of our math family!

  3. #2
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Quote Originally Posted by sfspitfire23 View Post
    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,
    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).
    Follow Math Help Forum on Facebook and Google+

  4. #3
    Super Member
    Joined
    Mar 2008
    Posts
    845
    Thanks
    3
    Awards
    1
    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.
    Follow Math Help Forum on Facebook and Google+

  5. #4
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    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
    Follow Math Help Forum on Facebook and Google+

  6. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    13,975
    Thanks
    99
    Awards
    1
    Quote Originally Posted by sfspitfire23 View Post
    [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?
    Follow Math Help Forum on Facebook and Google+

  7. #6
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    10,312
    Thanks
    38
    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:

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

    There are eleven different sums (2 through 12).



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

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

    There are nine different sums (3 through 11).

    Follow Math Help Forum on Facebook and Google+

  8. #7
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Quote Originally Posted by Plato View Post
    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.

    \binom{6}{2}-6
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Comb application 2
    Posted in the Discrete Math Forum
    Replies: 6
    Last Post: January 24th, 2011, 01:25 AM
  2. comb or perm?
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: August 5th, 2010, 07:16 AM
  3. Replies: 1
    Last Post: April 17th, 2009, 01:34 PM
  4. Perm/Combinations problem
    Posted in the Statistics Forum
    Replies: 4
    Last Post: January 19th, 2009, 05:46 AM
  5. Perm/Comb/Pascal Math/Geometry- Please help!
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: May 19th, 2008, 08:56 AM

/mathhelpforum @mathhelpforum