Question

l.a. wedding caterers offers a wedding reception buffet. suppose a menu is planned around three different salads, six entrees, four side dishes, and seven desserts. there are eight different choices of salads, ten different choices of entrees, eight different choices of side dishes, and ten different choices of desserts. how many menus are possible? 44800 × menus resources ebook

Explanation:

Step1: Calculate salad combinations

We need to choose 3 salads from 8. Use combination formula: $\binom{n}{k}=\frac{n!}{k!(n-k)!}$
$\binom{8}{3}=\frac{8!}{3!(8-3)!}=\frac{8\times7\times6}{3\times2\times1}=56$

Step2: Calculate entree combinations

Choose 6 entrees from 10.
$\binom{10}{6}=\frac{10!}{6!(10-6)!}=\frac{10\times9\times8\times7}{4\times3\times2\times1}=210$

Step3: Calculate side dish combinations

Choose 4 sides from 8.
$\binom{8}{4}=\frac{8!}{4!(8-4)!}=\frac{8\times7\times6\times5}{4\times3\times2\times1}=70$

Step4: Calculate dessert combinations

Choose 7 desserts from 10.
$\binom{10}{7}=\frac{10!}{7!(10-7)!}=\frac{10\times9\times8}{3\times2\times1}=120$

Step5: Multiply all combinations

Multiply the four results to get total menus.
$56\times210\times70\times120$
First compute $56\times210=11760$, then $70\times120=8400$, then $11760\times8400=98784000$

Answer:

98784000 menus