A certain law firm consists of 4 senior partners and 6 junior partners. How many different groups of 3 partners can be formed in which at least one member of the group is a senior partner? (Two groups are considered different if at least one group member is different.)
Answer and explanation please. Thanks in advance.
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papgust
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Here's my take,mpaudena wrote:A certain law firm consists of 4 senior partners and 6 junior partners. How many different groups of 3 partners can be formed in which at least one member of the group is a senior partner? (Two groups are considered different if at least one group member is different.)
Answer and explanation please. Thanks in advance.
Three scenarios would arise:
1. 1 SP & 2 JP OR
2. 2 SP & 1 JP OR
3. 3 SP & 0 JP
1. 4C1 * 6C2 = 19 * 3! (For diff arrangements inside the group) = 114
2. 4C2 * 6C1 = 36 * 3! = 216
3. 4C3 * 6C0 = 4 * 3! = 24
Therefore, 114+216+24= 354.
Can you let us know the OA? Hope my approach and answer are correct.
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sanjana
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Whenever I see the word atleast 1 I always go for the rule
P(a)=1-p(a')
In counting 1 would be the number of groups with no restriction,
Without restriction
Picking 3 from 10 ppl : 10c3 = 120
Picking 3 without any Senior partner : 6c3 = 20
No of ways with atleast 1 SP = 120-20 = 100.
P(a)=1-p(a')
In counting 1 would be the number of groups with no restriction,
Without restriction
Picking 3 from 10 ppl : 10c3 = 120
Picking 3 without any Senior partner : 6c3 = 20
No of ways with atleast 1 SP = 120-20 = 100.
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life is a test
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I also get 100, an alternative (longer!) way if it helps:sanjana wrote:Whenever I see the word atleast 1 I always go for the rule
P(a)=1-p(a')
In counting 1 would be the number of groups with no restriction,
Without restriction
Picking 3 from 10 ppl : 10c3 = 120
Picking 3 without any Senior partner : 6c3 = 20
No of ways with atleast 1 SP = 120-20 = 100.
there are 3 different ways the partners can be arranged: (1) 1S 2J, (2) 2S 1J, (3) 3S 0J
(1) is 4C1 * 6C2 = 4*15 = 60
(2) is 4C2 * 6C1 = 6*6 = 36
(3) is 4C3 = 4
(1) + (2) + (3) = 100
OA pls?
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sanjana
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Definitely not.. do try it out for some problems and u will see.papgust wrote:Sanjana,
I feel that this applies only for probability. Not sure whether it can be applied to a combinations prob.
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papgust
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I've almost done the same method, but unnecessarily multiplied by 3! with each scenario. Thought we should re-arrange 3 people within the group.life is a test wrote: I also get 100, an alternative (longer!) way if it helps:
there are 3 different ways the partners can be arranged: (1) 1S 2J, (2) 2S 1J, (3) 3S 0J
(1) is 4C1 * 6C2 = 4*15 = 60
(2) is 4C2 * 6C1 = 6*6 = 36
(3) is 4C3 = 4
(1) + (2) + (3) = 100
OA pls?
In combinatorics, when do we need to re-arrange like i did for this prob and when we must not? Pls help me understand
[/quote]
I've almost done the same method, but unnecessarily multiplied by 3! with each scenario. Thought we should re-arrange 3 people within the group.
In combinatorics, when do we need to re-arrange like i did for this prob and when we must not? Pls help me understand[/quote]
I think your confused about the parenthetical, which says that the groups are different if one member is different. Here the order of the groups is not a factor so you don't have to rearrange the three. The order would matter if the problem said something like: "each group is considered different than another depending on the order that the members are chosen." Hope that helps.
I've almost done the same method, but unnecessarily multiplied by 3! with each scenario. Thought we should re-arrange 3 people within the group.
In combinatorics, when do we need to re-arrange like i did for this prob and when we must not? Pls help me understand[/quote]
I think your confused about the parenthetical, which says that the groups are different if one member is different. Here the order of the groups is not a factor so you don't have to rearrange the three. The order would matter if the problem said something like: "each group is considered different than another depending on the order that the members are chosen." Hope that helps.
