Question

In how many different ways can the director of a research laboratory choose two chemists from among 9 applicants and three physicists from among 6 applicants?

Answer 1

Answer:

math_helper(2236) (Show Source): You can put this solution on YOUR website!

Chemists can be chosen in 7C2 ways

Physicists can be chosen in 9C3 ways

Together: (7C2)(9C3) = (21)(84) = highlight%281764%29

Answer 2

To find: the number of different ways does the director of a research laboratory choose two chemists from among 9 applicants and three physicists from among 6 applicants?

Solution:

Find the number of ways the director of research chooses two chemists from among 9 applicants.

$\[\begin{array}{l}{ = ^9}{C_2}\\ = \frac{{9!}}{{2!\left( {9 - 2} \right)!}}\end{array}\]$

$\[ = \frac{{9 \times 8 \times 7!}}{{\left( {2!} \right) \times \left( {7!} \right)}}\]$

$\[\begin{array}{l} = \frac{{9 \times 8}}{{2 \times 1}}\\ = 36\end{array}\]$

Find the number of ways the director of research chooses three physicists from among 6 applicants.

$\[\begin{array}{l}{ = ^6}{C_3}\\ = \frac{{6!}}{{3!\left( {6 - 3} \right)!}}\end{array}\]$

$\[\begin{array}{l} = \frac{{6 \times 5 \times 4 \times 3!}}{{\left( {3!} \right) \times \left( {3 \times 2 \times 1} \right)}}\\ = 20\end{array}\]$