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## Precalculus

### Course: Precalculus>Unit 8

Lesson 4: Combinations

# Combination example: 9 card hands

Thinking about how many ways we can construct a hand of 9 cards. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

• What about the suits? Do they have any relevance in this? Would we multiply the final result shown by 4? • and then if i do care about the order, i just don't add the extra k! in the denominator, right? • That's right. Notice that if you had k people out of order and you cared about ordering them, you could choose the first person in k ways, the second person in k-1 ways, and all the way down to the last person being the only person left, so it works out that there were k! ways to sort them.
• Say a word has multiple letters repeating- like ARCHITECTURE- the R, C,E and T repeat. The practice questions only have one letter repeating like PLOP (P) or ERROR (R). How would you compute the number of distinct permutations you can get from the word ARCHITECTURE? • You just need to divide by the number of combinations for each duplicate letter.

ARCHITECTURE has 12 letters with 2 Rs, 2 Cs, 2 Ts, and 2 Es so you have 12!/(2!*2!*2!*2!) = 29,937,600 permutations

Another example would be MISSISSIPPI, 11 letters with 4 Ss, 4 Is, and 2 Ps and 11!/(4!*4!*2!) = 34,650 arrangements.
• Why is it not 36P9? • in 36P9 we would count a hand if the same cards in different order. however whatever order the cards come in, if they are the same, it will be the same hand. we would not want to count the same hand twice. hence order wont matter. so, we use 36C9. ie. we divide 36P9 with the number of different ways 9 things can be ordered (9!). hope i am right.
• how can i know that a particular question is permutation or combination • In permutations, the order of the results matters.
- lottery number
- serial number
- phone number
- safe code

In combinations, the order does not matter.
- card games
- nominees for government office
- pizza toppings

In order to tell the difference, just ask yourself if the order of the results matters:
Yes? = permutation
No? = combination

Hope this helps!
• What if the denominator became 0 factorial? • Why does dividing 36! by 9! make sense? I still don't understand why this formula give us the combination?
Sorry, I always ask these silly questions... T_T • What Sal did can be written as: 36!/(27! * 9!)

Where:

- 36! is the number of ways 36 cards can be arranged.
- 27! is the number of ways the remaining 36 - 9 = 27 cards can be arranged. It makes sense, since you don't care about the arrangement of the cards you are not going to have in a 9-card hand.
- 9! is just the number of ways you can arrange your hand after picking the 9 cards. It makes sense, because if I gave you 9 cards, you can arrange them however you want (arrange them by color, by type of suit, correlatively, etc.), but all those arrangement you can make don't change the fact that they are still the same 9 cards I gave you before.

Permutations care about those arrangements you can make after receiving the cards (i.e, order matters), while combinations don't.
• This keeps coming up in the practice, and while I can answer it now due to repetition, I don't understand it. is there a formula (my brain likes formulas)? You need to put your reindeer, Jebediah, Lancer, Ezekiel, Gloopin, and Bloopin, in a single-file line to pull your sleigh. However, Lancer and Ezekiel are best friends, so you have to put them next to each other, or they won't fly.
How many ways can you arrange your reindeer? • Think of the "best friends" as one unit/reindeer, then figure out how many ways there are to arrange four reindeer in a single-file line.

Here's the trick: remember how we pretended the two "best friends" were one reindeer unit? In each of those arrangements, either Lancer or Ezekiel can come first! And this gives you twice as many arrangements as you figured out in the first step.

So, multiply the number of arrangements x2 to get the final total.

Hope this helps!
• how do I calculate the number unique combinations if we have 4 letters (A,B,C and D) where AB is equal BA (similar to handshaking) and each letter by itself is a combination. The following shows all the acceptable combinations:
A
AB
ABC
ABCD
AC
ACD
BC
BCD
BD
C
CD
D • One thing I wondered while doing the exercises is: why is n! / k! (n-k)! always resulting in an integer? I understand that it should in these problems but don't see how to prove it based on this formula. • Well C= nPr/r! which is C= n!/r!(n-r)!.
Remember that n>r by definition. if n=r the n answer is 1! . So we know that r! divises n! (if n>r, n!/r! gives you an integer ex: 5!/3!= 5*4).

We also know that (n-r)! divises n! (if n>(n-r), n!/(n-r)! gives you an integer ex: 5!/(5-2)!= 5!/3!= 5*4)
We are almost done, since we know than n! is divisble by r! AND (n-r)!. There is a property in number theory that stats the following: if b divises a and c divises a then bc divises a.
(EDIT: this is only true when no prime divises bc in its reduced term. E.g bc< or equal to a.
Which is exactly what we have in the case of the combinatorics, n!>r!(n-r)!, we can proof this by induction.)

Induction: n!>or equal r!(n-r)! for n>or equal r ---> true i.e. 1!>or equal 1!(1-1)! =1
(n+1)!>r!(n+1-r)! <=> n!(n+1)>r!(n-r)!(n+1-r).
We now have to prove : (n+1)>(n+1-r) which is the case for every n>or equal to r.
QED.

So we can say that n!/r!(n-r!) will always result to be an integer if n is greater or equal to r.