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I'm studying binomial expansion, and to get all the outcomes or arrangements of head appearing twice after throwing a coin 6 times without using a tree diagram, it's written that we can use 6c2 to get the answer.

But I totally don't understand how 6!/2!4! is the same thing as 6c2. I need order to matter, here, I don't want for instance THHTTTT to be the same as THTTTHT. And when I read 6c2 the only thing I can imagine is two empty spaces (_ _) and me trying to make combinations of 6 objects in them.

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If you choose two suitable places (for Heads) among six, write H to them, and write T to the other places, you'll obtain a combination. And, there are (6 choose 2) such choices. It's equivalent to permuting HHTTTT because any different permutation corresponds to a choice of two suitable places for the Heads.

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  • $\begingroup$ I still can't get it :( $\endgroup$
    – Manar
    Sep 25, 2020 at 21:50
  • $\begingroup$ Think about 3 tosses, and 1 Head. The situations are HTT, THT and TTH. When you fix a place for H, T's should sit in other available places. in other words, there aren't any two combinations having same places for H. When you fix H, it's finished. $\endgroup$
    – gunes
    Sep 25, 2020 at 21:54
  • $\begingroup$ I still don't understand it, even in permutations problems when I was faced with permutations that had empty spaces like, say, parking 3 identical cars in 7 parking places. I couldn't directly say 7c3, I had to imagine the other 4 empty parking places as ghost cars which are all identical and say 7!/3!•4! to be able to imagine the situation. I couldn't visualise 7c3 $\endgroup$
    – Manar
    Sep 25, 2020 at 22:29
  • $\begingroup$ ${7 \choose 3} = \frac{7!}{3!(7-3)!} = \frac{7!}{3!\cdot 4!}$ by definition. $\endgroup$
    – BruceET
    Sep 26, 2020 at 5:36
  • $\begingroup$ Yes I know the rule but I wanted to visualise it, like I do with permutation where I can imagine a tree diagram and for combination I can just imagine a bunch of thick branchs each thick branch contains smaller branches which all have same objects but in different orders, just same objects but moving around and changing seats/places, making n! arrangements. Or imagine a person picking some objects from a box and putting each on a definite number of titles, each object occupying a whole tile. $\endgroup$
    – Manar
    Sep 27, 2020 at 4:10

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