Problem Solving

Hi,

I have read the explanations given for this problem in other threads; but my question is based on understanding the slot method better - more specifically, how do we know which factorial to divide by..
"A certain law firm consists of 4 senior partners and 6 junior partners. How many different groups of 3 partners can be formd 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)"

NOw by the slot method:
1. 1 Sr + 2 Junior is given as (4*6*5)/2!
My question is why is it 2! and not 3! .. I know the rule is that we divide by the number of interchangable items.. but can someone help me understand this..
By NOT dividing by 3!, are we considering SJJ, JSJ, and JJS as 3 different groups?? I assume that 4 * 6 *5 contains ALL possible combinations/scramblings of (S, J, J).

wazzawayne wrote:Hi,

I have read the explanations given for this problem in other threads; but my question is based on understanding the slot method better - more specifically, how do we know which factorial to divide by..
"A certain law firm consists of 4 senior partners and 6 junior partners. How many different groups of 3 partners can be formd 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)"

NOw by the slot method:
1. 1 Sr + 2 Junior is given as (4*6*5)/2!
My question is why is it 2! and not 3! .. I know the rule is that we divide by the number of interchangable items.. but can someone help me understand this..
By NOT dividing by 3!, are we considering SJJ, JSJ, and JJS as 3 different groups?? I assume that 4 * 6 *5 contains ALL possible combinations/scramblings of (S, J, J).

we have 4 seniors and 6 Juniors
We want to form a 3 person committee that will comprise ATLEAST one senior member.


Regular Approach

Committees can formed as follows

1senior AND 2juniors
OR
2senior AND 1junior
OR
3seniors AND 0juniors
----------------------------
4C1 AND 6C2
OR
4C2 AND 6C1
OR
4C3 AND 6C0
----------------------------
4C1 X 6C2
+
4C2 X 6C1
+
4C3 X 6C0
----------------------------
4 X 15
+
6 X 6
+
4
----------------------------
60 + 36 + 4 -----------> 100


Shorcut

Committees can formed as follows

No of Committees with ATLEAST one senior = Total Possible committees - Committees that comprise NO Senior

Total Possible Committees = We know, in all, there are 10 members. A committee can be formed by selecting any 3 members among 10 members. This can be done in 10C3 ways.

Committees that comprise NO senior = This committee will comprise only juniors. This can be done in 6C3 ways.

So,
No of Committees with ATLEAST one senior = 10C3 - 6C3 = (10 X 9 X 8 / 3 X 2 X 1) - (6 X 5 X 4 / 3 X 2 X 1)
= 120 = 20 = 100

Regards,

Abhijit
PS :- Check my article https://gmatclub.com/forum/permutations- ... l#p1211661 on GMATCLUB for thru understanding of Permutations and Combinations concept. The same article will be posted on BEAT THE GMAT very soon. [/u]