FightToSurvive wrote:
Vyshak wrote:
FightToSurvive wrote:
A dance delegation of 4 people must be chosen from 5 pairs of dance partners. If 2 dance partners can never be together on the delegation, how many different ways are there to form the delegation?
A. 20
B. 30
C. 60
D. 70
E. 80
Bunnuel, from what I understood from the question stem is that 2 people who are dance partners cannot be in the delegation team of 4.
In total we have 10 people ( 5 pairs) out of which 2 dance partners cannot be on the team.
Hence to select a team of 4 people from 8. = 8C4 = 70. D IMO.
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A. 20
B. 30
C. 60
D. 70
E. 80
Bunnuel, from what I understood from the question stem is that 2 people who are dance partners cannot be in the delegation team of 4.
In total we have 10 people ( 5 pairs) out of which 2 dance partners cannot be on the team.
Hence to select a team of 4 people from 8. = 8C4 = 70. D IMO.
Posted from my mobile device
Hi FightToSurvive,
The method you have adopted is incorrect. You are just removing 2 members from the 10 members and then selecting a committee of 4 members. Out of 8 members, there is still a possibility of selecting members from the same pair.
Hope it helps.
I know that... I just thought that the question asked to remove any 2 people who are dance partners..
Anyways thanks for clarifying.
Could you please explain your approach to the solution?
Posted from my mobile device
Question: 2 dance partners can never be together on the delegation - It means we can select only one dancer from any given pair.
We have a total of 5 pairs of members. We have to select a delegation of 4 members. This can be done by choosing 4 members such that only one member will be picked up from one pair.
Out of 5 pairs, we will select 4 pairs. --> 5C4 ways
But each pair will have 2 dancers and hence there will be 2 ways of picking a member from each pair. So we have 4 pairs and we will get 2*2*2*2 = 16 ways of picking the members from the selected 4 pairs.
Hence, total number of ways = 5C4 * 2 * 2 * 2 * 2 = 80
Hope it helps.
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