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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!

Sagot :

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

[tex]5C2=\frac{5!}{(5-2)!\cdot2!}=\frac{120}{6\cdot2}=\frac{120}{12}=10[/tex]

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

[tex]6C3=\frac{6!}{(6-3)!\cdot3!}=\frac{720}{6\cdot6}=\frac{720}{36}=20[/tex]

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

[tex]4C1=\frac{4!}{(4-1)!\cdot1!}=\frac{24}{6\cdot1}=4[/tex]

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

[tex]3C2=\frac{3!}{(3-2)!\cdot2!}=\frac{6}{1\cdot2}=3[/tex]

The total possibilities are:

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

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