HELP! I thought I understood comb/perms but my entire thinking is stuck on this one concept - why doesnt the anagram or line method work for coin toss possibilities?
EX: If I flip a coin 3 times, and want to know how many possible outcomes there are, why can I not use the anagram, HHHTTT?
I know from listing the possibilities, or just multiplying 2*2*2, the number of possible combinations is 8. However, if I used the anagram above, it would be 6!/3!3! which is 20.
What am i missing here??
Coin toss possibilities - why doesn't anagram method work???
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- anshumishra
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Here are the 8 possible combinations you are looking for :
HHH
HHT
HTH
HTT
THH
THT
TTH
TTT
Thanks
HHH
HHT
HTH
HTT
THH
THT
TTH
TTT
Thanks
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Thanks, I know that there are 8 possibilities from listing them out. My problem is that I just don't see where the flaw lies in using the nCr formula for this one. I need to correct this so it doesn't mess me up elsewhere.
In theory, how would you use factorials to determine the number of possible outcomes from N tosses of a coin?
In theory, how would you use factorials to determine the number of possible outcomes from N tosses of a coin?
- anshumishra
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Actually, the two events (getting a head and getting a tail) are not independent.
What you are basically doing is calculating :
3C0*3C3+3C1*3C2+3C2*3C1+3C3*3C0 = 20
0-head,3 tails + 1 -head, 2-tails + 2-heads, 1-tail, + 3-heads, 0 -tail
But the fact is that, once you select :
0 head - > it automatically means 3 tails
1 head -> automatically means 2 tails
2 heads -> automatically means 1 tail
3 heads -> 0 automatically means 0 tail
So, the number of possible outcomes :
3C0+3C1+3C2+3C3 = 1+3+3+1 = 8
Hopefully, it helps !
Thanks
What you are basically doing is calculating :
3C0*3C3+3C1*3C2+3C2*3C1+3C3*3C0 = 20
0-head,3 tails + 1 -head, 2-tails + 2-heads, 1-tail, + 3-heads, 0 -tail
But the fact is that, once you select :
0 head - > it automatically means 3 tails
1 head -> automatically means 2 tails
2 heads -> automatically means 1 tail
3 heads -> 0 automatically means 0 tail
So, the number of possible outcomes :
3C0+3C1+3C2+3C3 = 1+3+3+1 = 8
Hopefully, it helps !
Thanks
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