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

Question

Ten nominees for three business awards are randomly seated at the 10 places at a round table. Three of these nominees are then called up together to receive the three awards. What is the probability that no two adjacent seats will be left empty?

A. 1/3
B. 2/5
C. 5/12
D. 7/16
E. 1/2

A. 1/3 
B. 2/5 
C. 5/12 
D. 7/16 
E. 1/2

ScottTargetTestPrep · Accepted Answer

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!. In between these 7 non-winners around the table, there are 7 spots between every pair of them. 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

Answer: C

IanStewart · Answer

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.

CareerGeek · Answer

Favourable outcome = No two adjacent person should win the award.

—> Let the winners be A, B and C
Let us arrange the remaining 7 persons in a circle = (7-1)! = 6!

Now, we have 7 spaces between these people, in which the 3 winners can be arranged in 7p3 ways

So, favourable ways = 6!*7p3

Total ways of arranging 10 people in a circle = 9!

Required Probability = 6!*7p3/9! = 7*6*5/(7*8*9) = 5/12

IMO Option C

Posted from my mobile device

nick1816 · Answer

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

prachilulla · Answer

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

nisen20 · Answer

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.
 https://gmatclub.com/forum/download/file.php?id=66645 
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 .

IanStewart · Answer

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.

nisen20 · Answer

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.

avggmatstudent23 · Answer

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?

IanStewart · Answer

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!

JROYENG97 · Answer

Solution using the formula for circular permutations, and without differentiating between the 7 losers and 3 winners:

Number of possible permutations of the 10 people, 7 losers and 3 winners, in a circular non symmetrical arrangement: (10-1)!/(7!3!) = 12

Note: The expression of the denominator is meant to account for the permutations of the 7 losers and 3 winners around the fixed circle considered in the numerator. (10-1)! Represents the 10! permutations of the 10 people, divided by the ten permutations that are identical rotations of one another (ref.: circular permutation formula).

Number of desirable permutations: Given 7 losers positioned around a circle, there are 7 spaces between them. The number of ways to position the 3 winners in these 7 positions: (7C3)/7 = 5

Note: Similarly to the concept for circular permutation formula, we divide by the number of available spaces here, 7, to account for arrangements which are identical rotations of one another.

.: Probability that no two adjacent seats will be left empty: 5/12

Answer: C.

KarishmaB · Answer

Rather than picking people out from a round table, let's think about what happens when they are going to sit at the table. We have 3 nominees and 7 regular people. 
We want that the 3 nominees should not sit together so that when they get up, there are no two adjacent empty seats. This becomes a regular circular arrangement problem:

7 R and 3 N are to be seated around a circular table such that no two N are seated together.

For this question assume that a person brings in his chair when he comes to sit. It is easy to visualize then. 
We first make all 7 Rs sit around the table in 6! ways (because it is circular arrangement)
Then we select 3 of the 7 spots between them in 7C3 ways and arrange the 3 Ns on those 3 spots in 7C3 ways.

Number of arrangements when no Ns are seated together = 6! * 7C3 * 3!
Total number of arrangements of 10 people sitting around a circular table = 9!

Required probability = 6! * 7C3 * 3! / 9! = 5/12

Answer (C)

shwetakoshija · Answer

To find  : P(no two adjacent seats empty) = 1 - P(2 or 3 adjacent seats empty)
 
 P(2 or 3 adjacent seats empty) = P(exactly 2 adjacent seats empty) + P(3 adjacent seats empty)

No. of ways of having 3 adjacent seats empty :
 
 Choose the first empty seat (Seat - 1) in 10C1 = 10 ways. 
 Choose the next two empty seats (Seats 2 and 3) (clockwise) in 1 way.   
So, no.(3 adjacent seats empty) = 10 x 1 x 1 = 10 ways.

No. of ways of having exactly 2 adjacent seats empty :

Choose the first empty seat (Seat - 1) in 10C1 = 10 ways. 
 Choose the next empty seat (Seat - 2) (clockwise) in 1 way. 
 Now, the third seat cannot be Seat - 10 or Seat - 3. So, choose the third seat in 6 ways.

So, no.(exactly 2 adjacent seats empty) = 10 x 1 x 6 = 60 ways.

Hence, P(2 or 3 adjacent seats empty) = (10 + 60)/10C3 = 7/12

Answer  : P(no two adjacent seats empty) = 1 - 7/12 =  5/12

QuantMadeEasy · Answer

Here's how I solved this question.

https://www.youtube.com/watch?v=tHgDdtjVNuo

Smart Prep (Tutor)

Wadree · Answer

Woah...*sighs.... dis one will be a big one....less go....
We wanna know probability of *no adjacent person is awarded* .... In other words... Everyone awarded needs to have at least 1 person between each other..... Now..... We will figure out da probability of da opposite.... We'll figure out da probability that.... At least 2 person among da 3 persons awarded... Are adjacent.....
Now... At least 2 are beside each other... This can happen in 3 ways....

1.
.............O.....O
.........O............O
.......O.................O
.........O.............P
..............O...O
First... One awarded person ( P ) sits anywhere of da 10 spots..... NOW REMAINING SPOTS = 9 
.............O.....O
.........O............O
.......O.................G2
.........O.............P
..............O...G1
Then.... another awarded guy sits on any of da 2 adjacent spots ( G1 and G2 ) ... beside ( P ) ..... So.... probability of another guy sitting on 2 spots among 9 spots is ...... 2 / 9......

2. 
.............O.....O
.........O............S2
.......O.................O
.........O.............P
..............S1..O
Now.... first awarded guy ( P ) .... and another awarded guy sits on exactly any of da 2 spots ( S1 and S2 ) .... look carefully... exactly any of these 2 spots... That are exactly 1 spot after ( P ) ...... So probability of someone sitting on any of 2 spots among 9 spots is..... 2 / 9 ......
LETS SAY ...DA SECOND GUY SITS ON ( S2 ) .... SO SPOTS REMAINING = 8

.............O.....O
.........O............S2
.......O.................O
.........O.............P
..............O...O

.... So.....TO MAKE AT LEAST 2 OF THE AWARDED GUYS ADJACENT... DA THIRD GUY NEEDS TO SIT ON ANY OF DA 3 SPOTS ( L1....L2....L3 ) ....

.............O.....L1
.........O............S2
.......O.................L2
.........O.............P
..............O...L3
So... probability of someone sitting on 3 spots among 8 spots is.... 3 / 8.....

So... probability of dis scenario is = ( 2 / 9 ) × ( 3 / 8 ) = 1 / 12

3.
Now...last scenario.....lets say one awarded guy ( P ).... Another awarded guy ( S ).... But ( S ) doesn't sit beside ( P ) .... Nor does he sit exactly 1 spot after ( P ) ....... LETS SAY ( S ) SITS ANYWHERE OF DA 5 SPOTS ( S1...S2...S3...S4...S5 ) .... AMONG 9 REMAINING SPOTS..... so probability of someone sitting on any of 5 spots among 9 spots is..... 5 / 9.....

.............S2.....S1
.........S3............O
.......S4.................O
.........S5.............P
..............O...O
Lets say....he sits on ( S3 ).... So spots remaining = 8 
.............O.....O
.........S3............O
.......O.................O
.........O.............P
..............O...O
SO...TO MAKE AT LEAST 2 OF THE AWARDED GUYS ADJACENT... THIRD GUY NEEDS TO SIT ON ANY OF DA 4 SPOTS .....
( L1...L2...L3...L4 ) .....

.............L1.....O
.........S3............O
.......L2.................L3
.........O.............P
..............O...L4

so... probability of someone sitting on 4 spots among 8 spots is.....4 / 8.......

So.... probability of dis scenario = ( 5 / 9 ) × ( 4 / 8 ) = 5 / 18

********FINALLY*********
TOTAL PROBABILITY OF AT LEAST 2 GUYS BESIDE EACH OTHER...... ( 2 / 9 ) + ( 1 / 12 ) + ( 5 / 18 ) = 7 / 12 ......

So.... probability of NO GUY beside each other = *TOTAL PROBABILITY* — (7 / 12 ) = 1 - ( 7 / 12 ) = 5 / 12

!nah id win!

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

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