Combinatorics

[mpaudena]

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.

[papgust]

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.

Here's my take,

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.

[sanjana]

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.

[papgust]

Sanjana,
I feel that this applies only for probability. Not sure whether it can be applied to a combinations prob.

[life is a test]

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.

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?

[sanjana]

Sanjana,
I feel that this applies only for probability. Not sure whether it can be applied to a combinations prob.

Definitely not.. do try it out for some problems and u will see.

[mpaudena]

Sorry for the delay with OA. I didn't get an email for some reason.

Any way, Sanjana is correct. It is 100.

[papgust]

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?

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

[mpaudena]

[/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.

[papgust]

Yup. I understand now. Thanks for clarifying!