Number of ways of choosing 3 distinct kinds of rolls for first guest
=(3/9*3/8*3/7)*(3!)
=9/28
Number of ways of choosing 3 distinct kinds of rolls for second guest
=(2/6*2/5*2/4)*3!
=2/5
Number of ways of choosing 3 distinct kinds of rolls for third guest
=(1/3*1/2*1/1)*3!
=1/1
Probability that all three guests get three distinct rolls
=9/28 * 2/5 * 1/1
=9/70
So 9/70 = m/n => m+n = 9+70 = 79
A hotel packed a breakfast for each of three guests. Each breakfast should have consisted of three types of rolls, one each of nut, cheese, and fruit rolls. The preparer wrapped each of the nine rolls, and, once they were wrapped, the rolls were indistinguishable from one another. She then randomly put three rolls in a bag for each of the guests. Given that the probability that each guest got one roll of each type is m/n, where m and n are relatively prime integers, find m+n.
1 answer
1 answer