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## Precalculus

### Course: Precalculus>Unit 8

Lesson 4: Combinations

# Handshaking combinations

Sal figures out how different combinations of people can shake hands.

## Want to join the conversation?

• When I paused to think about this I found that if you have N people and you follow the rules of this video then you will have an amount of handshakes that are equal to the sum of all positive integers that are less than N.
For example, with 8 people you will have `7+6+5+4+3+2+1 = 28` handshakes.

But you will also have `N! / ((N-2)! * 2)` handshakes.

So that means that the sum of all positive integers less than N is equal to `N! / ((N-2)! * 2)` when N is an integer that is greater than 1. Is this true? If so then why is this? • Also, you can think the handshake question like this:
Take 5 people for example.
Then A will shake hand with B, C, D, E.
B will shake hand with C, D, E (A is counted).
C will shake hand with D, E (A, B are counted).
D will shake hand with E
add them up you get 4 + 3 + 2 + 1 which is the sum you discovered.

Also, if you draw a pentagon and name the vertices A, B, C, D, E,
all the diagonals and sides of the pentagon would represent the handshakes:
AB, AC, AD, AE, BC, BD, BE, CD, CE, DE. Count them and it's easy to find
there are 10 lines.
• at , how can there be 4 possibilities, one person only have 3 options and not his own self. right? •  The 4 means that the first person involved in shaking hands can be any of the 4 people.
The 3 means that the second person involved in shaking hands can be any of the remaining 3 people not counting the person identified as the first person.
• Why is it that when you roll two die, there are 21 different combinations that can arise, if you don't care about the order of rolling the die? For example, your first die being 4 and your second die being 3 is the same thing as the first die being 3 and the second die being 4? •  Alright, let's reason through this.
Since, we only care about combinations, the order in which we roll the dice doesn't matter. So, yes. First die being 4 and the second die being 3 is the same thing as the first die being 3 and the second die being 4.
Now, for the combinations as such.
Let's list out all the ordered pairs first.
``(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)``

Remember these are the possibilities when we care about order and it is equal to 6*6 = 36.
Since for combination order doesn't matter, (1,4) = (4,1) and so on, we have a number of redundant ordered pairs in our list of possibilities. Let's remove them and at the same time count out the possibilities.
``(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)    Count = 6(2,2) (2,3) (2,4) (2,5) (2,6)    Count = 5(3,3) (3,4) (3,5) (3,6)    Count = 4(4,4) (4,5) (4,6)    Count = 3(5,5) (5,6)    Count = 2(6,6)    Count = 1``

So, all the possible combinations = 6+5+4+3+2+1 = 21. Hope this makes this a bit more clear.
• Is there any way to reason through the number of combinations without using permutations or the combination formula? • Can combination be explained in arithmetic series. In this case say 3+2+1
(A can shake with 3 other people, B with 2 other since B has already shaken hands with A and C with 1 and D with none)
which is 6 • Yes, but only for combinations in which you are choosing groups of 2, like the handshake problem. The formula for choosing 2 items out of n items is n!/(2! * (n-2)!) = n(n-1)/2, and as you correctly noticed, this is also the formula for the sum of the arithmetic series 1 + 2 + ... + (n-1) = n(n-1)/2.

The pattern does not continue to hold for combinations of 3 or more, however.
• How did ( 4 ) turn into 4*3/2
(2) • So, in other words, is combination a subset of permutation? • I'm having a difficult time grasping the difference in selection criteria. Were I choosing two teams out of a certain number of people, and each team had a name, this would be the permutation (criteria of 2) and if the teams had no names this would be the combination? Or do I have that backwards? If there are 12 teams and 4 must go to the 2nd category at the end of the season, the number of ways this can happen is ₁₂C₄ (are combinations. If team A, B, C and D went to the 2nd category, the order at which they do so doesn't matter).

If there 12 teams and 3 can win a medal (gold, silver or bronze) then the number of ways a medal can be won is ₁₂P₄ (are permutations. It's not the same having Team A with gold, B with silver and D with bronze than B with gold, D with silver and A with bronze. So order do matter).

Choosing between permutations or combinations has to do with the order, and the order has to do with those things in real life where we care about order and things we don't.

(:
• so.. i tried to solve this one yet i couldnt.. can anyone help?

How many different nine-digit numbers can be created by moving digits in the number 123454321

thanks • I need some help with this problem:
From 4 Mathematicians and 3 Statisticians, find the number of committees of size 3 that can be formed with 2 Mathematicians and 1 Statistician.

What I did is 4C2 + 3C1 = 9 which I did use sci cal for easier computation, however, is it correct? or do I have to multiply both?
4C2 x 3C1 = 18
Another one is 5 CBA, 4 CAS, and 3 CEIT students, a committee consisting of 3 CBA, 2 CAS, and 1 CEIT. How many ways can be done if:
a. any student from these colleges can be included.
5C3 + 4C2 + 3C1 = 19 or I have to multiple together?
5C3 x 4C2 x 3C1 = 180

When it comes to multiple entities with varying condition permutations and combinations really get confusing when various conditions are applied. I need tips. :) 