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06 Jul 2026, 09:54
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Solve this question using bi-junction, which is, total outcomes - unfavorable outcomes = Favorable outcomes
Total outcomes = 8C3 = 56 ways
Total unfavorable outcomes =
Case 1: when AB are considered = 2C2 * 6C1 = 6 ways (Explanation: when AB together are selected then it would be an unfavorable case and the remaining members would be 6: 1 way to select AB and 6 ways to select 1 out of 6 members)
Case 2: When CD are considered = 2C2 * 6C1 = 6 ways (Explanation: When CD together are selected then it would be unfavorable case and the remaining members would be 6: 1 way to select CD and 6 ways to select 1 out of 6 members)
Therefore using bi-junction =
56-6-6 = 44 ways = favorable outcomes
Probability = favorable outcomes/total outcomes = 44/56 = 11/14
Total outcomes = 8C3 = 56 ways
Total unfavorable outcomes =
Case 1: when AB are considered = 2C2 * 6C1 = 6 ways (Explanation: when AB together are selected then it would be an unfavorable case and the remaining members would be 6: 1 way to select AB and 6 ways to select 1 out of 6 members)
Case 2: When CD are considered = 2C2 * 6C1 = 6 ways (Explanation: When CD together are selected then it would be unfavorable case and the remaining members would be 6: 1 way to select CD and 6 ways to select 1 out of 6 members)
Therefore using bi-junction =
56-6-6 = 44 ways = favorable outcomes
Probability = favorable outcomes/total outcomes = 44/56 = 11/14
06 Jul 2026, 09:57
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Finding the total possibility of Committees: 8C3. = 56.
Committees that will disqualify to be valid
1. With Ally and Ben
So if the committee contains Aly and Ben: there will be 1 position open for the teams so 6C1 = 6 Teams in total that will not qualify.
2. Cam and Dan.
If the committee contains Can and Dan then there will be again 1 position open for the teams so 6C1 Teams will again not qualify.
So out of the total 56: 56- (6 + 6) will be valid Committees. = 44
The probabilty: 44/56.
= 11/14.
So IMO X = 11 and Y = 14.
Committees that will disqualify to be valid
1. With Ally and Ben
So if the committee contains Aly and Ben: there will be 1 position open for the teams so 6C1 = 6 Teams in total that will not qualify.
2. Cam and Dan.
If the committee contains Can and Dan then there will be again 1 position open for the teams so 6C1 Teams will again not qualify.
So out of the total 56: 56- (6 + 6) will be valid Committees. = 44
The probabilty: 44/56.
= 11/14.
So IMO X = 11 and Y = 14.
Bunuel
06 Jul 2026, 10:02
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Total ways to form a committee = 8C3 = 8!/(3!* 5!) = 56 ways
A committee is invalid if It contains
Both Ally & Ben or,
Both Cam & Dan
Ways when Aly & Ben in the committee,
6 C 1 = 6
Ways when Cam & Dan in the committee,
6 C 1 = 6
No overlap in the both consition as three member committee is to be formed.
Invalid committee = 6 + 6 = 12
Valid committee = 56 - 12 = 44
Probability to form valid committee,
= 44/ 56 = 11/14
Answer :
X = 11
Y = 14
A committee is invalid if It contains
Both Ally & Ben or,
Both Cam & Dan
Ways when Aly & Ben in the committee,
6 C 1 = 6
Ways when Cam & Dan in the committee,
6 C 1 = 6
No overlap in the both consition as three member committee is to be formed.
Invalid committee = 6 + 6 = 12
Valid committee = 56 - 12 = 44
Probability to form valid committee,
= 44/ 56 = 11/14
Answer :
X = 11
Y = 14
Bunuel
