Question

[Londondrug]

There are 8 chairs in a row. 8 persons,including 3 males and 5 females would take a seat there. What's the probability that at least 2 males will sit beside each other?

[Academic Actuary]

Do you have a numerical answer? I got 26/56 but I don't know if it is correct.

[daaaave]

I have a couple of arguments that give 36/56.

Serious suggestion for small combinatorics problems: list out all the possibilities and count. There are 8 choose 3 = 56 ways to seat the people, so listing them and counting should take you under 3 minutes. More importantly, listing them will help you see patterns that will let you understand where the combinatorial formulas come from. Not listing cases in a combinatorics problem that you can't do is like not drawing a picture in a geometry problem that you can't do.

[daaaave]

Four solutions:

1. List out the possibilities. Number the chairs 1 through 8 from left to right. Then to have at least 2 men beside each other, we need the men to sit in one of the following arrangements of seats:

Spoiler:

123 124 125 126 127 128
234 235 236 237 238
134 345 346 347 348
145 245 456 457 458
156 256 356 567 568
167 267 367 467 678
178 278 378 478 578

There are 36 such cases, giving us 36/56.

2. Keep in mind the list from the previous solution. How can we count them efficiently? Of the men who sit together, one of them is on the left.

Spoiler:

If the leftmost man sits in seat 1, then a second sits in 2, and there are 6 seats left for the 3rd guy.

If the leftmost man in the clump sits in seat n for n>=2, then the second sits in n+1, and the 3rd can sit anywhere but n, n+1 (since those are full) or n-1 (since that would give one of our other cases). So there are 5 places for the 3rd guy to sit. The leftmost man in the group can sit anywhere from 1 to 7. He can't be in seat 8 since no one is to the right, and we already did n=1, leaving 6 places for the first guy, for a total of 6*5 combinations here.

That gets us up to the same 6+6*5=36 combinations that worked from the list, or 36/56 for our total.

3. Counting differently: How many ways can all 3 sit together? How many ways can exactly 2 sit together?

Spoiler:

For all 3 to be together, the leftmost guy can be in any seat from 1 to 6, so there are 6 ways to do that.

For exactly 2 to be together, you could be together at one end (i.e., seats 1 and 2 or 7 and 8). To avoid having all 3 together, you need a woman to be next, leaving 5 seats for the 3rd man. There are 2 ends, so there are 2*5 ways for that to happen.

If they aren't together at one end, you have FMMF somewhere. The leftmost man in that group can be in any seat from 2 to 6, or 5 choices. There are 4 seats left for the 3rd man, or 5*4 choices left.

That gives a total of 6+2*5 + 5*4 =36.

4. We can count the complement: How many ways can the men sit all apart?

Spoiler:

To have the 3 men not sit together, the leftmost man has a woman on the right, for a MF somewhere. Likewise, the middle man has a woman on the right, for a second MF. There is no such constraint on the 3rd man as he might be in the rightmost seat, so he is just a lone M somewhere. There are 3 women left to be assigned, so we really have 6 "people" with 3 single F, our two MF, and our one M. There are 6 choose 3 ways to choose where the F sit, for 20 ways.

There are 56 total ways to choose which 3 seats the men get, 20 have the men apart, so 56-20=36 ways for at least 2 to be together.

[Academic Actuary]

Originally Posted by daaaave

I have a couple of arguments that give 36/56.

Serious suggestion for small combinatorics problems: list out all the possibilities and count. There are 8 choose 3 = 56 ways to seat the people, so listing them and counting should take you under 3 minutes. More importantly, listing them will help you see patterns that will let you understand where the combinatorial formulas come from. Not listing cases in a combinatorics problem that you can't do is like not drawing a picture in a geometry problem that you can't do.

I agree with 36/56. I forgot the cases with MMW in the first three seats and WMM in the last three.

[Londondrug]

Thanks Davvvve for your input and suggestions.
I meet with this kind of problem from ASM test occasionally, not very confident about it. It cost me much more time than other questions because I can't find a general solution to deal with it. Now I have your solutions.

to Academic Actuary, I created this question by my own, as you were, not so confident with my own answer and solution, so I put it down here.

Thanks again for doctor Daaave.