2.2.1 Binomial coefficients

The expression $\frac{n!}{r!(n-r)!}$ occurs as:

• •

  the number of permutations of $n$ objects composed of two types with $r$ of one type and $n-r$ of the other

• •

  the number of ways of selecting $r$ objects from $n$ distinguishable objects.

Binomial coefficients are written $\left({\displaystyle \genfrac{}{}{0pt}{}{n}{r}}\right)={\displaystyle \frac{n!}{r!(n-r)!}}$ for $r=0,1,\mathrm{\dots },n$ and $n=1,2,\mathrm{\dots }$. We read $\left(\genfrac{}{}{0pt}{}{n}{r}\right)$ as “$n$ choose $r$”.

Choosing three of the five coin tosses to show H can be done in $\left(\genfrac{}{}{0pt}{}{5}{3}\right)=10$ different ways. They are

HHHTT, HHTHT, HTHHT, THHHT, HHTTH,

HTHTH, THHTH, HTTHH, THTHH, TTHHH.

$\mathrm{\Omega }$ consists of all the sequences of H and T of length 5.
Each coin toss can be one of two outcomes, H or T.
So there are $2\times 2\times 2\times 2\times 2=2^5=32$ different sequences of H and T.
This is the size of the sample space $\mathrm{\Omega }$.

We have already seen that the event we care about has size $\left(\genfrac{}{}{0pt}{}{5}{3}\right)=10$.

So $\mathrm{P}(\text{three heads})=10/32.$