Answer to Question #140901 in Algebra for Cameron Foster

Question

1. A committee of six is chosen from ten girls and eight boys. How many committees can be formed if. . .
(a) There are no restrictions?
(b) The committee contains only girls?
(c) The committee contains only boys?
(d) The committee contains four girls?
(e) The committee contains at least three girls (Set up but don’t solve)



2. Compa has twelve lemon, ten vanilla, and fifteen chocolate puddings in her pouch. She hands Neptune
four random puddings with no replacement nor regard to order. How many batches of four can Compa hand Neptune if . . .
(a) there are no restrictions?
(b) Neptune receives only lemon puddings?
(c) Neptune receives two chocolate puddings?
(d) Neptune receives one chocolate, two lemon, and one vanilla pudding?
(e) Neptune receives at least two lemon puddings (set up but don’t solve)

Answer 1

1. Girls = 10 ; Boys = 8

a. No restrictions

$^{10}$C₆ × $^{8}$C₀ + $^{10}$C₅ × $^{8}$C₁ + $^{10}$C₄ × $^{8}$C₂ + $^{10}$C₃ × $^{8}$C₃ + $^{10}$C₂ × $^{8}$C₄ + $^{10}$C₅ × $^{8}$C₁ + $^{10}$C₀ × $^{8}$C₆

b. Contains Only Girls

$^{10}$C₆ × $^{8}$C₀

C. Contains only Boys

¹⁰C₀ × ⁸C₆

D. Contains 4 girls

¹⁰C₄ × ⁸C₂

E. Contains At least three girls

$^{10}$C₆ × $^{8}$C₀ + $^{10}$C₅ × $^{8}$C₁ + $^{10}$C₄ × $^{8}$C₂ + $^{10}$C₃ × $^{8}$C₃

2. Lemon = 12

Vanilla = 10

Chocolate = 15

a. No restrictions

¹²C₄ × ¹⁰C₀ × ¹⁵C₀ + ¹²C₀ × ¹⁰C₄ × ¹⁵C₀ + ¹²C₀ × ¹⁰C₀ × ¹⁵C₄ + ¹²C₃ × ¹⁰C₁ × ¹⁵C₀ + ¹²C₀ × ¹⁰C₃ × ¹⁵C₁ + ¹²C₁ × ¹⁰C₀ × ¹⁵C₃ + ¹²C₁ × ¹⁰C₃ × ¹⁵C₀ + ¹²C₀ × ¹⁰C₁ × ¹⁵C₃ + ¹²C₃ × ¹⁰C₀ × ¹⁵C₁ + ¹²C₂ × ¹⁰C₁ × ¹⁵C₁ + ¹²C₁ × ¹⁰C₂ × ¹⁵C₁ + ¹²C₁ × ¹⁰C₁ × ¹⁵C₂ + ¹²C₂ × ¹⁰C₂ × ¹⁵C₀ + ¹²C₀ × ¹⁰C₂ × ¹⁵C₂ + ¹²C₂ × ¹⁰C₀ × ¹⁵C₂

b. Only lemon puddings

¹²C₄ × ¹⁰C₀ × ¹⁵C₀

C. Two chocolate puddings

¹²C₁ × ¹⁰C₁ × ¹⁵C₂ + ¹²C₀ × ¹⁰C₂ × ¹⁵₂ + ¹²C₂ × ¹⁰C₀ × ¹⁵C₂

D. One chocolate, two lemon and one vanilla pudding

¹²C₂ × ¹⁰C₁ × ¹⁵C₁

E. At least two lemon puddings

¹²C₂ × ¹⁰C₁ × ¹⁵C₁ + ¹²C₂ × ¹⁰C₂ × ¹⁵C₀ + ¹²C₂ × ¹⁰C₀ × ¹⁵C₂ + ¹²C₃ × ¹⁰C₁ × ¹⁵C₀ + ¹²C₃ × ¹⁰C₀ × ¹⁵C₁ + ¹²C₄ × ¹⁰C₀ × ¹⁵C₀