Answer to Question #281732 in Discrete Mathematics for Neha

Question

Wandana is going to toss a coin eight times. In how many ways can she get five

heads and three tails

Answer 1

The objective have to choose 3 tail results from the 8 events.

if we take O for T and 1 for H, then this can be imagined as an 8 -digit binary number composed of 3 zeros and 5 ones.

And we have to find that in how many different ways can we arrange 3 zeros in such a number?

To choose 3 from 8 we reckon like so:

There are 8 possibilities for the first "tail" ; 7 for the second and 6 for third. And, in the end we do not care about the order of these assignments. The calculation goes like so:

(8 * 7 * 6) /(3 * 2)

The more standard way to express this is:

8 ! /(5 ! * 3 !)=56

Alternatively, we can use the combination formula to choose 5 spots to place heads from 8 contiguous locations, that is given by

$\left(\begin{array}{l}8 \\ 5\end{array}\right)=8 ! /(5 ! * 3 !)=56$ which gives the same answer.