Combinatorics 04

knight247 wrote:Of the three digit integers less than 300, how many have two digits equal to each other and the remaining digit different from the other two?
A 53
B 54
C 55
D 59
E 60

First, we are dealing with integers from 100 to 299, so there are 200 numbers altogether (ignoring any rules for now).

Of these 200 numbers, each one falls into one of 3 categories:
case a: no digits the same
case b: 2 digits the same
case c: all 3 digits the same

Strategy: Determine the number of 3-digit integers in cases a and c and subtract them from 200. This will tell us the number of 3-digit integers in case b.

case a: no digits the same
We'll use the Fundamental Counting Principle.
We'll begin with the most restrictive stage

Stage 1 select a hundreds digit
Stage 2 select a tens digit
Stage 3 select a units digit

Stage 1: can be accomplished in 2 ways (must be either 1 or 2)
Stage 2: can be accomplished in 9 ways (once a digit is selected for stage 1, there are 9 digits to choose from)
Stage 3: can be accomplished in 8 ways

Total number of 3-digit integers where no digits are the same = 2x9x8 = 144

case c: all 3 digits the same
There are 2 such numbers. They are 111 and 222

So, the number of 3-digit integers in case b = 200 - 144 - 2 = 54

Cheers,
Brent