Question

10. You and your date go to a restaurant where there are 5 meats, 6 vegetables, 4 types of bread, and 3 desserts to choose from. In how many ways could you select 2 meats, 3 vegetables, 1 bread, and 2 dessertsformula : nCr= n! / (nr) r!

Answer 1

You have to select 2 meats from 5 possible meats, that is, 5C2 =

`5C2=(5!)/((5-2)!\cdot2!)=(120)/(6\cdot2)=(120)/(12)=10`

You have to select 3 vegetables from 6 possible vegetables, that is, 6C3 =

`6C3=(6!)/((6-3)!\cdot3!)=(720)/(6\cdot6)=(720)/(36)=20`

You have to select 1 bread from 4 possible types of bread, that is, 4C1 =

`4C1=(4!)/((4-1)!\cdot1!)=(24)/(6\cdot1)=4`

You have to select 2 desserts from 3 possible desserts, that is, 3C2 =

`3C2=(3!)/((3-2)!\cdot2!)=(6)/(1\cdot2)=3`

The total possibilities are:

5C2*6C3*4C1*3C2 = 10*20*4*3 = 2400