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Lets say I am given 3 sets:

$S_1 = \{a_1, a_2, a_3,...,a_{10}\}\quad$ $S_2 = \{b_1, b_2, b_3,...,b_8\}\quad$ $S_3 = \{c_1, c_2, c_3,...,c_5\}$

I am unsure how to find the total number of combinations if I am supposed to choose 2 elements from $S_1$, 2 elements from $S_2$ and 1 element from any of the sets. I am also unable to choose the same element twice in any combination.

An example of a combination could be $\{a_1, a_3, b_3, b_8, c_2\}$ where there is always 5 elements total out of 23 total elements.

My original answer was incorrect.

Any ideas?

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You might want to look at some basic counting princples, mainly en.wikipedia.org/wiki/Rule_of_product and en.wikipedia.org/wiki/Combination –  Cain Mar 1 '13 at 14:53

3 Answers 3

up vote 3 down vote accepted
  • Choose two of $10$ from set $S_1$ ($10$-choose-$8$).
  • $\times$
  • Choose two of $8$ from set $S_2$ ($8$-choose-$2$).
  • $\times$
  • For the fifth element, we choose $1$ from the $23 - 4 = 19$ remaining elements, knowing $19$-choose-$1$ = $19$ choices.

$$\binom{10}{2}\times \binom{8}{2}\times [(23-4)]= \binom{10}{2}\times \binom{8}{2}\times (19) = \quad?$$

Recall: $$n-\text{choose}-k = \binom{n}{k} = \dfrac{n!}{k!(n-k)!}$$


Note: we need both "combinations": n-choose-k, or and the "rule of the product", to solve this problem. Both these tools need to become very familiar to you.

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Okay so the end isn't the remaining total (19) C 1? –  Addison Mar 1 '13 at 14:52
    
19 choose 1 (you've selected 4 of the 23 when choosing the first four elements). Note that $\binom{19}{1} = 19$ –  amWhy Mar 1 '13 at 14:56
    
One more thing. What if I wanted all combinations of the 5 elements chosen? Wouldn't I just multiply by (5!) ? –  Addison Mar 1 '13 at 15:52
    
...after choosing the 5 elements as described, if order matters, then you could compute all possible arrangements (permutations) of those five elements by computing $5!$. –  amWhy Mar 1 '13 at 16:07
    
Okay so, if {a1, a2, b1, b2, c1} is one set, and {a2, a1, b1, b2, c1} is a set as well, isn't that just (10 C 2 * 8 C 2 * 17 * 5!)? –  Addison Mar 1 '13 at 16:09

If you are supposed to choose 2 elements from $s_1$, 2 elements from $s_2$, and one element from any of the sets, you can at first choose 2 elements from $s_1$ in $\binom{10}{2}$ ways then you choose 2 elements from $s_2$ in $\binom{8}{2}$ ways . Now after choosing these four elements you are left with $23-4=19 $ elements from which you have to choose 1 element , which you can do in $\binom{19}{1}$ ways .

So the total no. of ways=$\binom{10}{2}.\binom{8}{2}.\binom{19}{1}$

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So you have $C^2_{10}$ ways of choosing 2 elements of $S_1$, $C^2_8$ ways of choosing 2 elements of $S_2$, and finally 10+8+5-4=19 ways of choosing the last element. Hence you have 45.28.19=23940 combinations.

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