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Combinatorics - why does order matter here

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Combinatorics - why does order matter here

by wazzawayne » Mon Apr 22, 2013 6:32 am
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).
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by Anju@Gurome » Mon Apr 22, 2013 8:40 am
wazzawayne 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)
Total number of groups with at least one senior = Total number of groups without any restriction - Total number of groups with no seniors

Total number of groups without any restriction = Number of ways to pick 3 people from (4 + 6) people = 10C3 = 10*9*8/3! = 120
Total number of groups with no seniors = Number of ways to pick 3 people from 6 people = 6C3 = 6*5*4/3! = 20

Total number of groups with at least one senior = 120 - 20 = 100

To answer your query,
wazzawayne wrote: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 interchangeable items.. but can someone help me understand this..
There are 3 slots.
For (1 senior, 2 junior) group,
  • One slot must be filled by a senior.
    As there are 4 seniors, the slot can be filled in 4 ways.

    The other two slots must be filled by 2 juniors.
    As there are 6 juniors, the 1st slot can be filled in 6 ways and the 2nd slot can be filled in 5 ways. Now, for any of these, the two juniors can arranged among them in 2! ways.
    So, effective number of selection of 2 juniors = 5*6/2!

    Hence, total number of such groups = 4*[(5*6)/2!] = 4*5*6/2!
So it is not like we are selecting 3 items.
We are actually selecting one from a group and two from another group.
If it was 3 from same group it'd have been 4*5*6/3!, but in this case it is actually [4/1!]*[5*6/2!]

Hope that helps.

Note: I'd suggest you to learn and understand the combination formulas. As I've shown, using the formulas life becomes a lot more easier. In fact, even if you use slot method, using formulas, number of (1 senior, 2 junior) groups = (Number of ways to select one senior from 4)*(Number of ways to select two juniors from 6) = [4C1]*[6C2] = 4*(5*6/2!) --> You don't have to worry about 2! or 3!
Anju Agarwal
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by abhijitlandge » Mon Apr 22, 2013 9:59 am
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]
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