Ten nominees for three business awards are randomly seated at the 10

Suppose there are 3 winners A_1, A_2 and A_3, and a,b and c persons are sitting in between them.


a+b+c=7, where a,b,c≥1
Number of ways that at least 1 person is sitting in between them.=6C2=15

Total number of ways
a+b+c=7, where a,b,c ≥0
Total number of ways= 9C2=36

Probability= 15/36=5/12

Round table so possible ways of arranging 10 people ; (10-1)! ; 9!
now given that all three nominees are called at same time so the table will have 7 people who can be arranged in (7-1)! ; 6! ways ; the 3 winners who have left the table so the spaces between 7 can be filled in 6!*7*6*5 ways ;
total (P) ; 6!*7*6*5/9! ; 6!*7*6*5/9*8*7*6! ; 6*5/9*8 ; 5/12 ways
IMO C

We can assume the 10 people sitting adjacent as 5 couples
since only 3 out of 10 people can be chosen, total number of outcomes is 10C3= 120

number of desired outcomes= 5C3*5= 10*5 (*5 because 5 couples)

Probability= 50/120= 5/12

This is an old GMATPrep problem, and there's no need for circular permutation formulas. You can pick the 3 award winners in 10C3 = (10)(9)(8)/3! = 120 ways. We can then discard the arrangements we don't want to count. There are 10 ways to pick three winners who are all in adjacent seats (because from any one of the ten people, picking the next two people clockwise around the table will produce a different selection of three adjacent people). Similarly, there are 10 ways to pick two people in adjacent seats, and when we pick two adjacent people, there will be 6 people left at the table who are not adjacent to either of those two, so there are 6*10 = 60 ways to pick two people in adjacent seats and a third in a non-adjacent seat. So there are 10 + 60 = 70 ways to pick three people from this table so that at least two are in adjacent seats, and 50 remaining ways to pick three people so that none are adjacent, and the answer is 50/120 = 5/12.

Sir Stewart, your method is great. But can you tell me what's wrong in my reasoning?

Step 1: Pick any seat as the 1st winner-seat and name this seat as #1. Then two adjacent seats—#10 and #2—would be unavailable to be chosen.

Step 2: Pick #3 as the 2nd winner-seat. Then #4 would be unavailable. Now, 5 available seats are left, and they're #5, #6, #7, #8 and #9.
Step 3: Pick #4 as the 2nd winner-seat. Then #3 and #5 would be unavailable. 4 available seats are left, and they're #6, #7, #8 and #9.
Step 4: Pick #5 as the 2nd winner-seat. Then #4 and #6 would be unavailable. 4 available seats are left, and they're #3, #7, #8 and #9.
Step 5: Pick #6 as the 2nd winner-seat. Then #5 and #7 would be unavailable. 4 available seats are left, and they're #3, #4, #8 and #9.
Step 6: Pick #7 as the 2nd winner-seat. Then #6 and #8 would be unavailable. 4 available seats are left, and they're #3, #4, #5 and #9.
Step 7: Pick #8 as the 2nd winner-seat. Then #7 and #9 would be unavailable. 4 available seats are left, and they're #3, #4, #5 and #6.
Step 8: Pick #9 as the 2nd winner-seat. Then #8 would be unavailable. 5 available seats are left, and they're #3, #4, #5, #6 and #7.

In my reasoning, there are 30 ways satisfying the condition which requires no two adjacent seats will be left empty.

There's nothing wrong with your reasoning! Once you pick the first winner, there are 9 ways to pick the second winner, and 8 ways to pick the third. So there are 9*8 ways to pick the remaining two winners, and as you worked out, 30 ways to pick those two winners so no two adjacent seats are empty, so the answer is 30/9*8 = 5/12.

If you thought your answer was wrong because you used 9C2 as your denominator (so if you counted the last two selections as if order did not matter) then you'd need to count your numerator as if order does not matter also. So you'd then need to notice that you counted every pair of winners twice if order does not matter (you counted, e.g., the selection #3 and #7 once in "Step 2" and once in "Step 6"), so you would then need to divide your count of 30 by 2, to get 15 selections where no two adjacent seats are empty. That still gives the right answer - you get 15/9C2 = 5/12. If you thought your answer was wrong because your denominator was something else altogether, then you probably didn't account for the fact that you already chose one winner when you fixed the first winner in seat #1.

IanStewart
I thought the denominator was 120 and the probability was 30/120.
Now, I know I wrongly paired the ways I counted with an unrelated total possibilities.

Thank you for your explanation.

Hi Ian

I tried a similar approach but I can't figure out where I'm going wrong.
Total ways to choose 10 people to sit around the table is 10!/7!3! = 120 ways

Now, when trying to count the ways 2 people sit together, there are 10 possibilities for the first person and then 2 for the second one (because that person can sit on either side of the first person and it will still be adjacent. Then there are 5 possibilities to choose the third person. This comes to 10x2x5 = 100.

Similarly, we can choose 3 people all sitting together in 10x2x1 = 20 ways. This I understand is wrong because any which ways there can only be a maximum of 10 people. I think my logic is right but I still don't understand the reason behind it.

CAN YOU PLEASE HELP ME UNDERSTAND WHERE I AM GOING WRONG?

Much appreciated!
Can you help me understand how this is wrong?

This solution is close to being right, but the issue is that you're double-counting a lot of the possibilities. Say we label the first few chairs A, B, C and D. When you count how many ways to pick two people in adjacent seats, if you say "there are 10 choices for the first person, 2 for the adjacent person", then you might choose B first, then A, or B first then C, for the two possibilities BA and BC. But your first choice might also be C, and then the second choice would be B or D, and we get the two possibilities CB and CD. But notice we already counted CB, because it's the same two people as BC, so we've counted that twice instead of once, and we'll end up doing that for every possibility.

We actually don't need to multiply anything to find out how many options we have to pick two people in adjacent seats -- if we think of picking two seats in a row, in clockwise order, we have 10 choices for the first seat, and then the adjacent seat, the one clockwise to the right of our first seat, is predetermined, so there are just 10 options in total (using the letters, they'd be AB, BC, CD, DE, etc until you get to JA). You then have 6 choices for the non-adjacent person (I think you used 5 in your calculation), for 60 possibilities in total. We similarly have just 10 choices to pick people in three adjacent seats (ABC, BCD, CDE, etc until we get to JAB).

It's a hard question - I hope that makes sense!

We need to find the probability that no two winners are sitting together.

Using the circular permutations formula (n - 1)!, we see that the total number of ways 10 people can sit around a table is (10 - 1)! = 9!.

The number of ways 7 non-winners can sit around a table is (7 - 1)! = 6!. [b]In between these 7 non-winners around the table, there are 7 spots between every pair of them.[/b] So the 3 winners have 7P3 ways to choose 3 of the 7 spots. Therefore, the number of ways that no two winners are sitting together is 7P3 x 6! (since each of them is separated non-winners).

Therefore, the probability that no two winners are sitting together is:

(7P3 x 6!)/9! = (7 x 6 x 5)/(9 x 8 x 7) = 30/72 = 5/12

ScottTargetTestPrep - If 3 winners left the 3 chairs, how are we getting 7 spots between every pair. Can you please help here as I'm unable to figure out this piece.

Answer: C