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06 Jul 2026, 08:00
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X: 11
Y: 14
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
64% (02:36) correct
36%
(02:58)
wrong
based on 515
sessions
History
Date
Time
Result
Not Attempted Yet
A company is forming a three-person leadership committee from a pool of eight qualified candidates: Aly, Ben, Cam, Dan, Emi, Fay, Gia, and Hal. Due to conflicts of interest, two pairs of candidates are incompatible: Aly and Ben cannot both serve on the committee together, and Cam and Dan cannot both serve on the committee together. A committee is considered valid only if its three members do not include either of these pairs of incompatible candidates.
Consider the following incomplete sentence:
If three candidates are chosen at random, the probability that the selected candidates form a valid committee is X / Y , where the fraction is expressed in simplest form.
Select for X and Y values that are consistent with the information provided. Make only two selections, one in each column.
Consider the following incomplete sentence:
If three candidates are chosen at random, the probability that the selected candidates form a valid committee is X / Y , where the fraction is expressed in simplest form.
Select for X and Y values that are consistent with the information provided. Make only two selections, one in each column.
| X | Y | |
| 3 | ||
| 7 | ||
| 11 | ||
| 13 | ||
| 14 |
ShowHide Answer
Official Answer
X: 11
Y: 14
|
⚠️ Important: GMAT Club does not allow AI-generated posts. AI-generated solutions are not eligible for kudos, and users who post them may face moderation action, including a ban.
06 Jul 2026, 09:00
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Manhattan Prep Official Explanation
A company is forming a three-person leadership committee from a pool of eight qualified candidates: Aly, Ben, Cam, Dan, Emi, Fay, Gia, and Hal. Due to conflicts of interest, two pairs of candidates are incompatible: Aly and Ben cannot both serve on the committee together, and Cam and Dan cannot both serve on the committee together. A committee is considered valid only if its three members do not include either of these pairs of incompatible candidates.
Consider the following incomplete sentence:
If three candidates are chosen at random, the probability that the selected candidates form a valid committee is X / Y , where the fraction is expressed in simplest form.
Select for X and Y values that are consistent with the information provided. Make only two selections, one in each column.
Step 1: Understand the Prompt and Question
The prompt describes a committee selection process with two incompatibility constraints and asks for the probability that a randomly chosen committee is valid. Because we need to count outcomes in order to find a probability, this is a Probability and Combinatorics problem.
The key formula is:
P(event) = Favorable Outcomes / Total Outcomes
To find this probability, we need to count the total number of possible committees and the number of valid ones — those that include neither forbidden pair. Rather than counting valid committees directly, it may be easier to count the invalid ones and subtract from the total.
Step 2: Plan your Approach
Use the Anagram Method to count the total number of possible committees: assign each candidate a letter — Y (selected) or N (not selected) — and count the distinct combinations using the formula \(\frac{?????!}{?!?!}\), where Y and N are the number of candidates assigned each of those letters, respectively. To calculate the number of valid committees, count the number of committees that contain a forbidden pair and subtract from the total number of possible committees.
Step 3: Solve the Problem
To find the total number of committees possible that select 3 people from 8 candidates, count arrangements of a word with 3 Y’s and 5 N’s:
Total = 8!/(5!3!) = 56
Next, find the number of invalid committees by counting how many committees contain a forbidden pair. If a committee contains Aly and Ben, there is only one more spot to be filled and six possible candidates to fill that spot, so there are a total of six committees that are invalid because of the Aly/Ben pair. The same logic applies for Cam and Dan: there are another six committees that are invalid because of the Cam/Dan pair. Out of 56 possible committees, 6 + 6 = 12 are invalid, leaving 56 - 12 = 44 valid committees. To compute the probability a randomly-selected committee is valid:
P(valid) = # Valid Committees / # Total Possible Committees
P(valid) = 44 / 56 = 11 / 14
The correct answer is 11 for the first column and 14 for the second column.
GMAT-Club-Forum-rqu9s0ub.png [ 8.36 KiB | Viewed 2002 times ]
A company is forming a three-person leadership committee from a pool of eight qualified candidates: Aly, Ben, Cam, Dan, Emi, Fay, Gia, and Hal. Due to conflicts of interest, two pairs of candidates are incompatible: Aly and Ben cannot both serve on the committee together, and Cam and Dan cannot both serve on the committee together. A committee is considered valid only if its three members do not include either of these pairs of incompatible candidates.
Consider the following incomplete sentence:
If three candidates are chosen at random, the probability that the selected candidates form a valid committee is X / Y , where the fraction is expressed in simplest form.
Select for X and Y values that are consistent with the information provided. Make only two selections, one in each column.
Step 1: Understand the Prompt and Question
The prompt describes a committee selection process with two incompatibility constraints and asks for the probability that a randomly chosen committee is valid. Because we need to count outcomes in order to find a probability, this is a Probability and Combinatorics problem.
The key formula is:
P(event) = Favorable Outcomes / Total Outcomes
To find this probability, we need to count the total number of possible committees and the number of valid ones — those that include neither forbidden pair. Rather than counting valid committees directly, it may be easier to count the invalid ones and subtract from the total.
Step 2: Plan your Approach
Use the Anagram Method to count the total number of possible committees: assign each candidate a letter — Y (selected) or N (not selected) — and count the distinct combinations using the formula \(\frac{?????!}{?!?!}\), where Y and N are the number of candidates assigned each of those letters, respectively. To calculate the number of valid committees, count the number of committees that contain a forbidden pair and subtract from the total number of possible committees.
Step 3: Solve the Problem
To find the total number of committees possible that select 3 people from 8 candidates, count arrangements of a word with 3 Y’s and 5 N’s:
Total = 8!/(5!3!) = 56
Next, find the number of invalid committees by counting how many committees contain a forbidden pair. If a committee contains Aly and Ben, there is only one more spot to be filled and six possible candidates to fill that spot, so there are a total of six committees that are invalid because of the Aly/Ben pair. The same logic applies for Cam and Dan: there are another six committees that are invalid because of the Cam/Dan pair. Out of 56 possible committees, 6 + 6 = 12 are invalid, leaving 56 - 12 = 44 valid committees. To compute the probability a randomly-selected committee is valid:
P(valid) = # Valid Committees / # Total Possible Committees
P(valid) = 44 / 56 = 11 / 14
The correct answer is 11 for the first column and 14 for the second column.
Attachment:
GMAT-Club-Forum-rqu9s0ub.png [ 8.36 KiB | Viewed 2002 times ]
General Discussion
06 Jul 2026, 08:15
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Reading the problem, invalid cases look much simpler than valid ones and Probability of Valid = 1- invalid, so calculating invalid cases first.
Invalid = When either AB or CD are present together.
case 1: If AB together, 1 place left for other 6 therefore 6 cases.
and same for case 2: CD together.
as we only need 3 members, both the cases cannot happen simultaneously. therefore, invalid cases = 6+6 = 12
total cases = 8C3 = 56
1- Prob(valid) = Prob(Invalid) = 12/56 = 3/14
therefore, Prob(valid) = 11/14
Invalid = When either AB or CD are present together.
case 1: If AB together, 1 place left for other 6 therefore 6 cases.
and same for case 2: CD together.
as we only need 3 members, both the cases cannot happen simultaneously. therefore, invalid cases = 6+6 = 12
total cases = 8C3 = 56
1- Prob(valid) = Prob(Invalid) = 12/56 = 3/14
therefore, Prob(valid) = 11/14
06 Jul 2026, 08:21
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This was my approach
Selecting A and B in the team - _A_ _B_ _6C1_
= 6C1
similarly selecting C and D into the team = 6C1
total ways 8C3
therefore X = 8C3 - 6C1 - 6C1
Y = 8C3
X/Y = 1 - {2(6)/(8*7*6/3*2*1)}
= 1 - 3/14 = 11/14
Selecting A and B in the team - _A_ _B_ _6C1_
= 6C1
similarly selecting C and D into the team = 6C1
total ways 8C3
therefore X = 8C3 - 6C1 - 6C1
Y = 8C3
X/Y = 1 - {2(6)/(8*7*6/3*2*1)}
= 1 - 3/14 = 11/14
