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GMAT Club World Cup 2026 (Day 1): A company is forming a three-person

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geocircle

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Three-person leadership committee from a pool of eight qualified candidates = 8C3 = 56
If a committee has Aly and Ben then the other member can be any of the other 6.
If a committee has Cam and Dan then the other member can be any of the other 6.

56-6-6 = 44 valid committees.

44/56 = 11/14

X=11 and Y=14

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Total ways to organize = 8c3 = 56
If AB are selected together, there will be 6 invalid committees, i.e., ABC, ABD, ABE, ABF, ABG, ABH
If CD are selected together, there will be 6 invalid committees, i.e., ACD, BCD, CDE, CDF, CDG, CDH

Therefore, 56 - 6 - 6 = 44 valid committees.
x = 44 = 11
y = 56 = 14
Answer, x/y = 11/14

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Probability = Valid/Total
Total = 8C3 = 56
Novalid = 6 (Aly, Ben, X), X has 6 possible values + 6 (Cam, Dan, X), X has 6 possible values = 12
Valid = Total - Novalid = 56 - 12 = 44

Probability = 44 / 56 = 11 / 14

X is 11
Y is 14

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ABCDEFGH ->8
The good committees probabilities is complimentary to the bad ones, so I will calculate the bad ones and subtract those probabilities from 1.
the probability of picking A is 1/8 and then the probability of picking B is 1/7. You can Pick A first and then B and vice versa, so as the probability of picking A is independent of the one of picking B the combined probability is their multiple - 2 * 1/8 * 1/7 = 1/28
Another bad committee is picking C and then D or vice versa, which is the same probability as above, so we get 1/28.

The bad committees will either be of the first or the second type, so we add the probabilities to get 1/28 + 1/28 = 2/28 = 1/14.
1- 1/14=13/14

Bunuel

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.

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Bunuel

A to H gives 8 candidates.

For favourable cases, 2 approaches are valid
First, considering valid cases or subtracting invalid cases from valid ones

Following the 2nd approach; all cases 8C3. Invalid cases are cases where AB_ or CD_ is a team the empty spots for each group can be filled by anyone of the rest 6 people so total 12 groups are not valid

Therefore answer is 8C3-12/8C3 = 44/56 => 11/14

X = 11 and Y = 14

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Aly, Ben, Cam, Dan, Emi, Fay, Gia, Hal

Incompatible: 12
Aly and Ben and a third one -> total 6
Cam and Dan and a third one -> total 6

Probability = (8C3 - 12) / 8C3 = (56-12)/56 = 44/56 = 11/14

X = 11
Y = 14

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Total ways to choose 3 member from 8 members = 8C3 ==> 56
Case where AB are chosen along with one other member = 1x1x6C1 ==>6
Case where CD are chosen along with one other member = 1x1x6C1 ==>6

We need to reduce both cases from total number of cases =56-(6+6) ==>44
Total valid number of cases ==> 44
Needed porbability = 44/56 ==> 11/14

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Bunuel

Total committees = 8C3 = 56

Invalid committees:

Aly and Ben together: 6 choices for the third person
Cam and Dan together: 6 choices for the third person

Total invalid = 6 + 6 = 12

Valid committees = 56 - 12 = 44

Probability = 44/56 = 11/14

X = 11; Y = 14

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X=11
Y= 14

Solution-
And : Valid Committees= Total committees - No of Invalid committees

Total Committees = 8C3 =56
Invalid :
Committees with Aly & Ben= 1 way to choose the pair, 6 ways to choose the third member. Hence 6 ways.
Committees with Cam and Dan= 1 way to choose the pair, 6 ways to choose the third member.
Hence 6 ways.
So Total Invalid= 6+6=12

Hence valid committees= 56-12=44
Therefore Ratio= 44/56 ie 11/14
Hence X= 11, Y= 14.

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Total ways to choose a committee of 3 from 8 people : 8C3 = 56

Now, count the invalid committees:

1. Both Aly & Ben are selected: The 3rd member can be any 1 of the 6 remaining candidates.
No. of such committees = 6
2. Both Cam & Dan are selected: Similarly, the 3rd member can be any 1 of the remaining 6 candidates.
No. of such committees = 6

There can be no overlap between the above 2 cases since it's only a 3 person committee.

Total invalid committees : 6+6 = 12
Valid committees : 56-12 = 44

Probability ( X/Y ) = 44/56 or 11/14

Answer : X = 11 and Y = 14

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Ans: X= 11, Y = 14

probability of an event happening will mean, {combination of all outcomes in which required event occurs/combination of all possible outcomes}

i) combination of all possible outcomes:
we'll use combinations to calculate all possible ways in which we can get 3 member teams from the 8 choices.using box & fill method: 8*7*6/3*2*1 =56

ii) combination of all outcomes in which required event occurs or what is not required does not occur:
we don't want an event in which A, B are together or C,D are together. so instead of going the long way about it we can use the mutual exclusive formula. so we'll first calculate scenarios in which we get A,B together on the team or C,D together on the team. for that we'll have to use "must be chosen together" method in combinations. then we'll subtract that from all possible outcomes to get to desired outcomes.

so 6 for A,B and 6 for C,D.
subtract 12 from 56 and you get the numerator.

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total possib= 8c3= 56

invalid committees include both A &B or both C&D
if both A&B chosen-> choose 3rd from 6 left =6c1=6
similarly if C&D chosen -> remaining 1 chosen from 6= 6c1=6

invalid committee= 6+6=12

valid=56-12=44

prob=44/56=11/14
x=11 y=14

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Two pairs - A/B and C/D cannot be in the committee. Note - Individual people from these pairs can be in the committee though, just not together.

TO = 8C3 = 56

C1 = A B __, third person can be chosen in 6C1 ways.

C2 = C D __, third person can be chosen in 6C1 ways.

Total = 12 selections are not valid.

FO = 56 - 12 = 44

P = 44/ 56 = 11/14

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Total outcome = 8C3 = 56

Favourable outcome = Total outcome - outcome where both are together

Ally & Ben = 6C1 = 6
Cam & dan = 6C1 = 6

Since there is no overlap, 6+6 = 12 which is to be deducted from 54. Therefore favourable outcome = 44

44/54 = 11/14

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Correct Answer: X = 11, Y = 14
Whenever a combinatorics question introduces "incompatible" or "cannot be together" rules, it is almost always faster to calculate the Total combinations and subtract the Invalid ones.

The Breakdown
1. Find the Total Possible Committees
We are choosing 3 people from a total pool of 8.

Total = 56 possible committees.

2. Find the Invalid Committees
A committee is invalid if it contains {Aly, Ben} OR {Cam, Dan}.

If Aly and Ben are chosen: They take up 2 of the 3 spots. There is only 1 spot left, which must be filled by one of the remaining 6 candidates. So, there are 6 invalid committees containing Aly and Ben.

If Cam and Dan are chosen: Following the exact same logic, they take up 2 spots, leaving 1 spot for the 6 remaining candidates. There are 6 invalid committees containing Cam and Dan.

(Note: We don't have to worry about double-counting a committee that has Aly, Ben, Cam, and Dan, because a committee only has 3 spots!)

Total Invalid Committees = 6+6=12.

3. Calculate the Probability

Valid Committees = Total − Invalid = 56−12=44.

Probability =
Total
Valid

=
56
44

.

Simplify the fraction by dividing the top and bottom by 4:

56
44

=
14
11

Therefore, X = 11 and Y = 14.

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Total number of ways the committee can be formed = 8C3 = 56
There are two cases where the invalid committee can be formed:
1. A and B are together and the third person is from the remaining 6 = 2C2*6C1 = 6
2. C and D are together and the third person is from the remaining 6 = 2C2*6C1 = 6
Total = 12

Probability(valid) = 1 - Probability(invalid)
= 1 - (12/56) = 1 - (3/14) = 11/14

Hence, X=11, Y =14

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Here is my solution. Please see the picture. You have three places to fill, so you choose 3 out of 8, "which is 8 over 3"(binomial coefficient), giving 56 possibilities overall. However, there are two restrictions: A and B cannot be together, and C and D cannot be together, which creates 12 invalid combinations. In the end, we calculate the probability, which number of valid combinations divided by the total number of possible combinations. Then we simplify the fraction and get 11/14.

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07 Jul 2026, 03:37

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x=3 and y=7

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All possible committees are 8C3 = 56
Committees in which the two banned pairs are included.
For AB WE 1X1X6= 6
For CD we get 1x1x6=6
Total unwanted cases= 6+6= 12
Valid committees=56-12 = 44
X/Y= 44/56
X= 11
Y=14

Bunuel

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So, the question asks about the probability of forming a valid committee if we choose 3 random candidates out of 8

P (E)= Total favorable combinations/Total combinations

Total combinations = 8C3

All possible cases for favorable outcomes:

1. We select one from each pair= 2C1 * 2C1 * 4C1 = 16
(One from each pair and one from the remaining four to select 3 valid candidates)

2. Select any one person from any one pair = 2C1 * 2C1 * 4C2 = 2*2*4 = 24
(First, select one pair out of the two pairs, then select one candidate out from the selected pair, and then from the remaining four, we select 2 candidates)

3. No candidates from either pair- 4C3- 4
We select 3 candidates out of E,F,G,H only

So, total favorable= 16+24+4 = 44

Probability= 44/56= 11/14