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.

Question

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.

Answer 1

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

Answer 2

Well, isn't this a roll-ercoaster of a problem! Let's break it down, shall we?

Since the rolls are indistinguishable once wrapped, we can consider them as a set of three rolls: nut, cheese, and fruit. Now, when we put three rolls in a bag for each guest, how many different combinations can we have?

For the first guest, they can receive any combination of the three rolls, so there are 3! = 6 possible combinations.

Similarly, for the second guest, there are 6 possible combinations.

And for the third guest, again, there are 6 possible combinations.

Now, to find the total number of outcomes, we need to multiply these possibilities together. So, 6 x 6 x 6 = 216.

But wait a minute! We're not done yet. Out of these 216 outcomes, how many of them result in each guest getting one roll of each type?

Well, let's think about it. The first guest can receive their rolls in any order, so that's 3 possible combinations.

The same goes for the second guest, 3 possible combinations.

And for the third guest, once again, 3 possible combinations.

Now we multiply these possibilities together: 3 x 3 x 3 = 27.

So, the probability that each guest gets one roll of each type is 27/216.

And since 27 and 216 have no common factors (except for 1), our answer is 27 + 216 = 243.

Hope that tickled your funny bone while solving the problem!

Answer 3

To solve this problem, we can use combinatorics. Let's consider the possible outcomes and favorable outcomes.

There are 9 rolls in total, and each guest gets 3 rolls. So the total number of ways to distribute the rolls is C(9,3) * C(6,3) * C(3,3), which represents choosing 3 rolls from the 9 for the first guest, then choosing 3 rolls from the remaining 6 for the second guest, and finally choosing the remaining 3 rolls for the third guest.

Now, let's consider the favorable outcomes where each guest gets one roll of each type. For the first guest, there are 3 rolls to choose from for each type, so there are 3^3 = 27 favorable outcomes. Similarly, for the second guest, there are 2 rolls of each type left, so there are 2^3 = 8 favorable outcomes. Finally, for the third guest, there is only 1 roll of each type left, so there is only 1^3 = 1 favorable outcome.

Thus, the total number of favorable outcomes is 27 * 8 * 1 = 216.

The probability that each guest gets one roll of each type is the ratio of the favorable outcomes to the total outcomes: 216 / (C(9,3) * C(6,3) * C(3,3)).

Calculating this probability and simplifying the fraction, we get:

216 / (84 * 20 * 1) = 9 / 35

Therefore, m = 9 and n = 35.

The sum m + n = 9 + 35 = 44.

Hence, the answer is 44.

Answer 4

To solve this problem, we can use the principle of counting known as permutations.

First, let's consider the total number of ways the preparer can distribute the nine rolls among the three guests. Since each roll is indistinguishable from one another once wrapped, we can think of this as arranging 9 identical objects into 3 distinct groups. This can be represented by stars and bars, where the stars represent the rolls and the bars separate the groups.

Using stars and bars, we need to place 2 bars among the 9 stars. There are (9+2) choose 2 ways to do this, which is equal to 11 choose 2 = 55.

Now, let's consider how many of these arrangements result in each guest receiving one roll of each type.

First, we can think about the number of ways to distribute the nut rolls among the guests. Each guest should get one nut roll, so this is equivalent to distributing 3 identical nut rolls among 3 distinct groups. Using stars and bars again, there are (3+2) choose 2 = 10 choose 2 = 45 ways to do this.

Similarly, the number of ways to distribute the cheese rolls among the guests is also 45, and the number of ways to distribute the fruit rolls is also 45.

To find the total number of arrangements where each guest receives one roll of each type, we multiply these three numbers together: 45 * 45 * 45 = 91125.

Finally, we can calculate the probability m/n. The probability that each guest gets one roll of each type is the number of successful arrangements (91125) divided by the total number of possible arrangements (55):

m/n = 91125/55

To find m+n, we simply add the numerator and denominator:

m+n = 91125 + 55 = 91180.

Therefore, the value of m+n is 91180.