Hi

#1 Direct counting:

one 2 + two 2's + three 2's = 1*5*5*3!/2! + 1*1*5*3!/2! + 1*1*1 = 75 + 15 + 1 = 91

# Indirect counting:

Total # of outcomes - # of outcomes without 2's = \(6^3 - 5^3 = 91\)

Answer C.

We need to determine, when rolling three dice, the number of outcomes in which at least one die shows a 2.

We should first recall the following formula:

Total number of ways to roll 3 dice = number of ways with at least one 2 + number of ways with no 2s.

Since there are 6 x 6 x 6 = 216 possible outcomes when rolling 3 dice, we have:

216 = number of ways with at least one 2 + number of ways with no 2s

number of ways with at least one 2 = 216 - number of ways with no 2s

There are 5 x 5 x 5 = 125 outcomes in which no 2 is shown.

Thus, there are 216 - 125 = 91 outcomes in which at least one 2 is shown.

Answer: C
_________________

Posted from my mobile device

Why do we divide by 2! In the direct counting method?