tanzzt wrote:
Hi Bunuel,
First of all, thanks for your reply.
Sorry that I'm still confused with the probability that it turns up heads exactly twice.
Does \frac{3!}{2!} mean flipping a coin three times & turning up exactly two heads?
So, if we have another question flipping a coin five times & turning up exactly three heads, it should be \frac{5!}{3!}. is that right?
Thanks.
Two heads and one tail can occur in three ways: HHT, HTH, THH. So basically the number of ways this particular scenario can happen is the number of permutations of three letters HHT, where two H's are identical.
If it were 3 heads and 2 tails, then the number of ways this scenario can happen would be 5!/(3!2!) = 10:
HHHTT;
HHTHT;
HTHHT;
THHHT;
THHTH;
THTHH;
TTHHH;
HTTHH;
HHTTH;
HTHTH.
Again, this is the number of permutations of 5 letters HHHTT, out of which 3 H's and 2 T's are identical.
THEORY:Permutations of \(n\) things of which \(P_1\) are alike of one kind, \(P_2\) are alike of second kind, \(P_3\) are alike of third kind ... \(P_r\) are alike of \(r_{th}\) kind such that: \(P_1+P_2+P_3+..+P_r=n\) is:
\(\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}\).
For example number of permutation of the letters of the word "gmatclub" is 8! as there are 8 DISTINCT letters in this word.
Number of permutation of the letters of the word "google" is \(\frac{6!}{2!2!}\), as there are 6 letters out of which "g" and "o" are represented twice.
Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be \(\frac{9!}{4!3!2!}\).
Hope it's clear.
P.S. For more follow the links in my previous post.
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