Question

A carton contains 12 light bulbs of which 4 are defective. In how many ways can you choose 4 of these bulbs and get at least two of the defective light bulbs?

Answer 1

Explanation:

so, we have actually 2 groups

12 - 4 = 8 working light bulbs.

4 defective light bulbs.

in our problem we actually check in how many ways we can pick 2 bulbs from the group of working bulbs AND 2 bulbs from the group of defective bulbs. or 1 and 3. or 0 and 4.

the sequence does matter in my understanding of the problem, as e.g.

W1W2D1D2 is a different selection than

W2W1D1D2

and so on.

that is permutations without repetitions (the same bulb cannot be picked multiple times in the same selection) :

P(n, k)

we have therefore

P(8, 2) = 8!/(8-2)! = 8!/6! = 8×7 = 56

P(8, 1) = 8!/(8-1)! = 8!/7! = 8

P(8, 0) = 8!/(8-0)! = 8!/8! = 1

and

P(4, 2) = 4!/(4-2)! = 4!/2! = 4×3 = 12

P(4, 3) = 4!/(4-3)! = 4!/1! = 4×3×2 = 24

P(4, 4) = 4!/(4-4)! = 4!/0! = 4×3×2 = 24

and the total possibilities of selections are

for 2 and 2

56 × 12 = 672

for 1 and 3

8 × 24 = 192

for 0 and 4

1 × 24 = 24

so, the total possibilities are

672 + 192 + 24 = 888