Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Hello my question is quite simple i would think but i just cant seem to find an answer. I have a set of $\{1,2,3,4,5,6,7,8,9,10\}$ and i would like to calculate how many unique given sets of $6$ can i get from this set. In other words for the number $1$ i would end up with $[1,2,3,4,5,6] [1,3,4,5,6,7] [1,4,5,6,7,8] [1,5,6,7,8,9] [1,6,7,8,9,10]$ I would move down the line with the number $2$ to compare to unique sets of $6$ note: when moving to two I would no longer do this $[2,1,3,4,5,6]$ because it repeats my first case above. its there a formula to figure this sort of thing? Thanks in advance.

when I work this out on paper i end up with 15 sets here is how

for 1
     [1,2,3,4,5,6]
    [1,3,4,5,6,7]
    [1,4,5,6,7,8]
    [1,5,6,7,8,9]
    [1,6,7,8,9,10]

  for 2
        [2,3,4,5,6,7]
        [2,4,5,6,7,8]
        [2,5,6,7,8,9]
        [2,6,7,8,9,10]
    for 3
            [3,4,5,6,7,8]
            [3,4,6,7,8,9]
            [3,5,6,7,8,9,10]
for 4 [4,5,6,7,8,9] 
      [4,6,7,8,9,10]
for 5 [5,6,7,8,9,10]

after that i cant make any more groups of $6$ thus i end up with $15$ sets.

share|improve this question
    
I'd be pretty certain there are other questions similar to this on the site that might help you. –  Simon Hayward Dec 2 '12 at 20:30
add comment

3 Answers

up vote 1 down vote accepted

Yes, there is, it is called the binomial, written $\binom{n}{k}$, read $n$ choose $k$. The value is $$\binom{n}{k}=\frac{n!}{k!(n-k)!}.$$ So, in your case, you have $$\binom{10}{6}=\frac{10!}{6!4!}=210.$$ I hope you find this helpful!

share|improve this answer
    
Forgive my stupidity Clayton I have two problems here. 1 how are we ending up with 210? the "!" operator is what exactly? Also i worked this out on paper and i only end up with 15 sets of 6 so I'm obviously missing something can you help me understand. –  Miguel Dec 2 '12 at 19:19
    
The "!" stands for the factorial, that is, $n!=n\cdot(n-1)\cdot \cdots\cdot (2)(1)$. So, $$10!=10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1.$$ Dividing by $6!4!$ is what gives you the 210 unique subsets of order 6. Does this make it more clear? –  Clayton Dec 2 '12 at 19:23
    
it makes it more clear but i just dont see how the resul makes sense. For example i edit my question can you look how i reached 15. Yet this formula is given me 210 im obviously missing something but i dont see what it could be? forgive my ignorance. –  Miguel Dec 2 '12 at 19:28
    
No worries, it is great to ask questions. Now, looking at the subsets you've listed, where would the set $[1,2,3,4,5,7]$ fit into? Also, we can have the sets $[1,2,3,4,5,8]$, $[1,2,3,4,5,9]$, and $[1,2,3,4,5,10]$, for example. Once we've exhausted one "component," which I'm taking to be the last one, change the next index over, so now we count $[1,2,3,4,6,7]$. So on and so forth. Hope it helps! –  Clayton Dec 2 '12 at 19:36
    
THANK YOU! :-D i knew there was something i was missing. Thank you! I truly appreciate your response. –  Miguel Dec 2 '12 at 19:40
add comment

Binomial coefficients count the number of distinct subsets of $k$ elements from a set containing $n$ elements. The notation for this is $\binom{n}{k}$ which is equal to $\frac{n!}{k!(n-k)!}$

share|improve this answer
add comment

It exactly the number of ways to choose $6$ elements out of $10$, i,e. the binomial coefficient$$\binom{10}{6}=\frac{10!}{6!4!}$$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.