The organizer of a conference is selecting workshops to include. She will select from 6 workshops about chemistry and 7 workshops about biology. In how many ways can she select 4 workshops if 2 or fewer must be about chemistry?

Respuesta :

Given that there are 6 workshops about chemistry and 7 workshops about biology.

So the total number of workshops available are,

[tex]\begin{gathered} =6+7 \\ =13 \end{gathered}[/tex]

The number of ways of selecting 'r' objects from 'n' distinct objects is given by,

[tex]^nC_r=\frac{n!}{r!\cdot(n-r)!}[/tex]

The total number of ways of selecting 4 workshops having no workshop about chemistry is calculated as,

[tex]\begin{gathered} n(\text{ 0 chemistry)}=^7C_4 \\ n(\text{ 0 chemistry)}=\frac{7!}{4!\cdot(7-4)!} \\ n(\text{ 0 chemistry)}=\frac{7\cdot6\cdot5\cdot4!}{4!\cdot3!} \\ n(\text{ 0 chemistry)}=\frac{7\cdot6\cdot5}{3\cdot2\cdot1} \\ n(\text{ 0 chemistry)}=35 \end{gathered}[/tex]

The total number of ways of selecting 4 workshops having exactly 1 workshop about chemistry is calculated as,

[tex]\begin{gathered} n(\text{ 1 chemistry)}=^7C_3\cdot^6C_1 \\ n(\text{ 1 chemistry)}=\frac{7!}{3!\cdot(7-3)!}\cdot\frac{6!}{1!\cdot(6-1)!} \\ n(\text{ 1 chemistry)}=\frac{7\cdot6\cdot5\cdot4\cdot3!}{3!\cdot4!}\cdot\frac{6\cdot5!}{1!\cdot5!} \\ n(\text{ 1 chemistry)}=\frac{7\cdot6\cdot5\cdot4}{4\cdot3\cdot2\cdot1}\cdot6 \\ n(\text{ 1 chemistry)}=210 \end{gathered}[/tex]

The total number of ways of selecting 4 workshops having exactly 2 workshops about chemistry is calculated as,

[tex]\begin{gathered} n(\text{ 2 chemistry)}=^7C_2\cdot^6C_2 \\ n(\text{ 2 chemistry)}=\frac{7!}{2!\cdot(7-2)!}\cdot\frac{6!}{2!\cdot(6-2)!} \\ n(\text{ 2 chemistry)}=\frac{7\cdot6\cdot5!}{2!\cdot5!}\cdot\frac{6\cdot5\cdot4!}{2!\cdot4!} \\ n(\text{ 2 chemistry)}=\frac{7\cdot6}{2\cdot1}\cdot\frac{6\cdot5}{2\cdot1} \\ n(\text{ 2 chemistry)}=315 \end{gathered}[/tex]

Consider that the number of ways to select 4 workshops if 2 or fewer must be about chemistry, will be equal to the sum of the individual cases when the number of chemistry workshops in the selection are either 0 or 1 or 2.

This can be calculated as follows,

[tex]\begin{gathered} \text{ Total}=n(\text{ 0 chemistry)}+n(\text{ 1 chemistry)}+n(\text{ 2 chemistry)} \\ \text{Total}=35+210+315 \\ \text{Total}=560 \end{gathered}[/tex]

Thus, the total number of ways is 560.