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

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

  • $\begingroup$ I'd be pretty certain there are other questions similar to this on the site that might help you. $\endgroup$ – Simon Hayward Dec 2 '12 at 20:30

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!

  • $\begingroup$ 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. $\endgroup$ – Miguel Dec 2 '12 at 19:19
  • $\begingroup$ 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? $\endgroup$ – Clayton Dec 2 '12 at 19:23
  • $\begingroup$ 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. $\endgroup$ – Miguel Dec 2 '12 at 19:28
  • $\begingroup$ 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! $\endgroup$ – Clayton Dec 2 '12 at 19:36
  • $\begingroup$ THANK YOU! :-D i knew there was something i was missing. Thank you! I truly appreciate your response. $\endgroup$ – Miguel Dec 2 '12 at 19:40

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


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 $\frac{n!}{k!(n-k)!}$


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