Difficult Selection Problem

[rickyishere]

Hi,

Can the quants out here help in solving this problem:

A firm has 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. ( 2 groups are considered different if at least one group member is different).

a) 48 , b) 100, c) 120 , d) 288, e) 600.

I ended up getting an answer of 4C1*6C2 + 4C2*6C1 but it does not equate to any of the answers above.

Thanks
Ricky

[thephoenix]

at least means >=1---> total- none

sme times using the other form helps much

we will find total combination and substract from that the total combination when none senior partners are selected

total partner=10(6+4)
we need to select =3
total combination=10C3=120

now total combination when none senior partners are selected
which means all three are from junior partners=6C3=20

reqd combination=120-20=100

[thephoenix]

Hi,

Can the quants out here help in solving this problem:

A firm has 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. ( 2 groups are considered different if at least one group member is different).

a) 48 , b) 100, c) 120 , d) 288, e) 600.

I ended up getting an answer of 4C1*6C2 + 4C2*6C1 but it does not equate to any of the answers above.

Thanks
Ricky

in your approach u missed for the case when all three partner are senior=4C3=4
its actually [4c1*6C2+6C1*4C2+4C3]
adding this to your method gives 100

hth

[harsh.champ]

Hi,

Can the quants out here help in solving this problem:

A firm has 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. ( 2 groups are considered different if at least one group member is different).

a) 48 , b) 100, c) 120 , d) 288, e) 600.

I ended up getting an answer of 4C1*6C2 + 4C2*6C1 but it does not equate to any of the answers above.

Thanks
Ricky

The simplest solution is as follows:-
atleast one member of the group is a senior partner=total no. of selections- when all members are junior partners
= 10C3 - 6C3 = 120 - 20 = 100

Hey ricky ,you had solved like this:-

I ended up getting an answer of 4C1*6C2[1 senior,2juniors] + 4C2*6C1[2 seniors,1junior] = 60 + 36 = 96 but it does not equate to any of the answers above.

But you left 1 case over here:- [3 seniors,no junior] = 4C3 = 4

So,now you see adding 4 to your upper answer ,you get 100.B which is the answer.

Hope,you don't have any doubts now.

[rickyishere]

Hi,

Can the quants out here help in solving this problem:

A firm has 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. ( 2 groups are considered different if at least one group member is different).

a) 48 , b) 100, c) 120 , d) 288, e) 600.

I ended up getting an answer of 4C1*6C2 + 4C2*6C1 but it does not equate to any of the answers above.

Thanks
Ricky

The simplest solution is as follows:-
atleast one member of the group is a senior partner=total no. of selections- when all members are junior partners
= 10C3 - 6C3 = 120 - 20 = 100

Hey ricky ,you had solved like this:-

I ended up getting an answer of 4C1*6C2[1 senior,2juniors] + 4C2*6C1[2 seniors,1junior] = 60 + 36 = 96 but it does not equate to any of the answers above.

But you left 1 case over here:- [3 seniors,no junior] = 4C3 = 4

So,now you see adding 4 to your upper answer ,you get 100.B which is the answer.

Hope,you don't have any doubts now.

I did not use the final combination because if I had selected 4C3 it would have meant that we are selecting all 3 seniors. However, choosing all 3 seniors can't be possible as per the problem because " 2 groups are considered different if at least one group member is different ". Thoughts on this?

[harsh.champ]

Hi,

Can the quants out here help in solving this problem:

A firm has 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. ( 2 groups are considered different if at least one group member is different).

a) 48 , b) 100, c) 120 , d) 288, e) 600.

I ended up getting an answer of 4C1*6C2 + 4C2*6C1 but it does not equate to any of the answers above.

Thanks
Ricky

The simplest solution is as follows:-
atleast one member of the group is a senior partner=total no. of selections- when all members are junior partners
= 10C3 - 6C3 = 120 - 20 = 100

Hey ricky ,you had solved like this:-

I ended up getting an answer of 4C1*6C2[1 senior,2juniors] + 4C2*6C1[2 seniors,1junior] = 60 + 36 = 96 but it does not equate to any of the answers above.

But you left 1 case over here:- [3 seniors,no junior] = 4C3 = 4

So,now you see adding 4 to your upper answer ,you get 100.B which is the answer.

Hope,you don't have any doubts now.

I did not use the final combination because if I had selected 4C3 it would have meant that we are selecting all 3 seniors. However, choosing all 3 seniors can't be possible as per the problem because " 2 groups are considered different if at least one group member is different ". Thoughts on this?

Well Ok,I am writing down those 4 selections over here.[In all of them atleast 1 group member are different,so we have 4 different cases]
Suppose the seniors are S1,S2,S3,S4.

4 possible selections:-
1) (S1,S2,S3)
2) (S1,S2,S4)
3) (S1,S3,S4)
4) (S2,S3,S4)

I guess this clears your doubt.

But I still don't understand why you didn't take the 4th case.
According to the above bold-faced statement of yours also,it is correct to take these 4 cases as they are all different.

[rickyishere]

Hi,

Can the quants out here help in solving this problem:

A firm has 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. ( 2 groups are considered different if at least one group member is different).

a) 48 , b) 100, c) 120 , d) 288, e) 600.

I ended up getting an answer of 4C1*6C2 + 4C2*6C1 but it does not equate to any of the answers above.

Thanks
Ricky

The simplest solution is as follows:-
atleast one member of the group is a senior partner=total no. of selections- when all members are junior partners
= 10C3 - 6C3 = 120 - 20 = 100

Hey ricky ,you had solved like this:-

I ended up getting an answer of 4C1*6C2[1 senior,2juniors] + 4C2*6C1[2 seniors,1junior] = 60 + 36 = 96 but it does not equate to any of the answers above.

But you left 1 case over here:- [3 seniors,no junior] = 4C3 = 4

So,now you see adding 4 to your upper answer ,you get 100.B which is the answer.

Hope,you don't have any doubts now.

I did not use the final combination because if I had selected 4C3 it would have meant that we are selecting all 3 seniors. However, choosing all 3 seniors can't be possible as per the problem because " 2 groups are considered different if at least one group member is different ". Thoughts on this?

Well Ok,I am writing down those 4 selections over here.[In all of them atleast 1 group member are different,so we have 4 different cases]
Suppose the seniors are S1,S2,S3,S4.

4 possible selections:-
1) (S1,S2,S3)
2) (S1,S2,S4)
3) (S1,S3,S4)
4) (S2,S3,S4)

I guess this clears your doubt.

But I still don't understand why you didn't take the 4th case.
According to the above bold-faced statement of yours also,it is correct to take these 4 cases as they are all different.

I did not choose the fourth case because to me " 2 groups are considered different if at least one group member is different " meant I need to have a senior and junior in the selection. I think this problem needs to be more specific because that last comment can be interpreted differently, well at least in my case.

[harsh.champ]

Hi,

Can the quants out here help in solving this problem:

A firm has 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. ( 2 groups are considered different if at least one group member is different).

a) 48 , b) 100, c) 120 , d) 288, e) 600.

I ended up getting an answer of 4C1*6C2 + 4C2*6C1 but it does not equate to any of the answers above.

Thanks
Ricky

The simplest solution is as follows:-
atleast one member of the group is a senior partner=total no. of selections- when all members are junior partners
= 10C3 - 6C3 = 120 - 20 = 100

Hey ricky ,you had solved like this:-

I ended up getting an answer of 4C1*6C2[1 senior,2juniors] + 4C2*6C1[2 seniors,1junior] = 60 + 36 = 96 but it does not equate to any of the answers above.

But you left 1 case over here:- [3 seniors,no junior] = 4C3 = 4

So,now you see adding 4 to your upper answer ,you get 100.B which is the answer.

Hope,you don't have any doubts now.

I did not use the final combination because if I had selected 4C3 it would have meant that we are selecting all 3 seniors. However, choosing all 3 seniors can't be possible as per the problem because " 2 groups are considered different if at least one group member is different ". Thoughts on this?

Well Ok,I am writing down those 4 selections over here.[In all of them atleast 1 group member are different,so we have 4 different cases]
Suppose the seniors are S1,S2,S3,S4.

4 possible selections:-
1) (S1,S2,S3)
2) (S1,S2,S4)
3) (S1,S3,S4)
4) (S2,S3,S4)

I guess this clears your doubt.

But I still don't understand why you didn't take the 4th case.
According to the above bold-faced statement of yours also,it is correct to take these 4 cases as they are all different.

I did not choose the fourth case because to me " 2 groups are considered different if at least one group member is different " meant I need to have a senior and junior in the selection. I think this problem needs to be more specific because that last comment can be interpreted differently, well at least in my case.

Well,usually we do consider the case when no junior is there.Unless otherwise given,we should not take the case that both the seniors and the juniors should be present.
In case of further query,I guess you should do "Word Translations" from Manhattan.
It is a very good guide and you will be more confident in formulating any word problem in the form of an equation.
Hope it helps