A firm has 4 senior partners and 6 junior partners. How many different

What seems to work as well, though its tougher to come up with this solution:

(10*9*8) * 2/3 + (10*9*4)*1/3 = 600. Divided by all possible permutations (=3!) yields 100.

Explanation: For the first spot you have 10 possible canditates, for the 2nd 9. For the 3rd spot you need to differentiate if a senior partner has been picked for one of the first 2 spots or not. If yes, then you can pick one of the 8 remaining guys, so 10*9*8. If no senior partner has been picked yet, you need 10*9*4.
Now you need to weight the 2 cases with the corresponding probabilities: probability of no senior partner being picked in the first 2 draws = 6/10*5/9=1/3, so you weight 10*9*4 with 1/3. For the complementary case (senior partner was picked in the first 2 draws) you just take the complementary prob (1-1/3)= 2/3 and weight 10*9*8 with it.
Now you just need to divide the answer (600) by the number of different positions (=3!) and get 600/6=100

But I suggest you stick with the easy solution in the GMAT

The standard approach

1 senior partner
4 x 6C2 = 60

2 senior partners
4C2 x 6 = 36

3 senior partners
4C3 = 4

Total 100
However GDKs approach is a better way around

A general approach to solving Combination problems

Steps:

1. There are two larger groups, the senior partners and the junior partners.
2. The first larger group i.e., the senior partners is 4 in number. So let\(n1\) be 4. The second larger group, ie, the junior partners is 6 in number. So \(n2\) is 6.
3. The smaller group that is selected from the larger group of senior partners may be 1, 2, or 3 in number. So \(r1\) is 1 or 2 or 3 . Correspondingly the other smaller group i.e.,\(r2\) selected from the junior partners is 2 or 1 or 0 in number.
4. For each value of\(r1\) and the corresponding\(r2\), compute the number of combinations which are\(4C1 * 6C2\), \(4C2 * 6C1\) and\(4C3 * 6C0\) being 60, 36 and 4 ways respectively.
5. The total number of combinations is therefore 60+36+4 = 100

You can also think about it as a code.

4S - 6J

three different ways to group them:

S J J. 4(6)5/2! because we have two junior members
or
S S J. 4(3)6/2! because we have two senior members
or
S S S 4(3)2/3! because we have three senior members.

add up the results and you get 100.

Hi All,

This is a rare layered Combinatorics question. To answer it, you can either figure out the Total of ALL possibilities and subject the ones that "don't fit" or you can perform 3 separate calculations to account for the ones that "do fit." Here is how you can approach the latter option:

We're given 4 senior partners and 6 junior partners. We're asked for the number of different groups of 3 (the clue that we'll need the Combination Formula) with one stipulation - there must be AT LEAST 1 senior partner. Here are the 3 calculations:

1) 3 senior partners = 4c3 = 4!/[3!1!] = 4 different groups

2) 2 seniors and 1 junior = (4c2)(6c1) = (4!/[2!2!])(6!/[1!5!) = (6)(6) = 36 different groups

3) 1 senior and 2 juniors = (4c1)(6c2) = (4!/[3!1!])(6!/[2!4!]) = (4)(15) = 60 different groups

4 + 36 + 60 = 100 different groups

GMAT assassins aren't born, they're made,
Rich

Hi,

Can some one help me understand what is wrong with my logic here?

A group should have 3 partners with at least 1 senior partner.
Overall there are 4 senior partners & 6 junior partners.

At least 1 senior partner = 4C1
Remaining 2 can be anyone from the remaining partners = (10-1)C2

4C1*9C2 = 144?

Can some one help me understand what i am missing here?

Hi ashwinchivukula,

Unfortunately, your math includes some 'duplicate entries.'

For example, let's call the 4 senior partners A, B, C and D and the 6 junior partners 1, 2, 3, 4, 5 and 6.

In your calculation, you state that the first person selected MUST be one of those 4 seniors (A/B/C/D) and the remaining two people can be any two of the remaining 9...

The group "A/B/1" and "B/A/1" are the SAME group, but your calculation counts THAT group TWICE (depending on whether A or B was chosen first). In a Combination question, you can't allow duplicate entries.

GMAT assassins aren't born, they're made,
Rich

Hi Bunuel

The question says - "How many different groups of 3 partners can be formed "
So we can calculate as:

Case - I
One group of 3Sp + 1Jp + second group of 1Sp + 2Jp

Case - II
One Group of 2Sp + 2Jp + Second group of 2Sp + 2Jp

Case -III
One Group of 2Sp + 2Jp + Second group of 1Sp + 2Jp + Third Group of 1Sp + 2Jp

Case - IV
One Group of 1Sp + 2Jp + Second group of 1Sp + 2Jp + Third Group of 1Sp + 2Jp

As per the ans - Its how many way a group of 3 Partners can be formed.
Or I seem to misinterpret the prompt.

The group must have 3 partners in it out of which at least one member is a senior partner:
3SP
2SP + 1JP
1SP + 2JP.

We are asked to find the number of groups with at least one senior partner.

“At least 1” means "one or more," so the group must have 1 or 2 or 3 senior partners.

Case 1: Exactly 1 senior partner

Recall that the group must have 3 partners. Therefore, in this case, we need to pick 1 senior partner from 4 senior partners and 2 junior partners from 6 junior partners. The number of ways this can be done is 4C1 x 6C2.

4C1 x 6C2 = 4 x (6x5)/2! = 4 x 15 = 60

Case 2: Exactly 2 senior partners

In this case, we need to pick 2 senior partners from 4 senior partners and 1 junior partner from 6 junior partners. The number of ways this can be done is 4C2 x 6C1.

4C2 x 6C1 = (4x3)/2! x 6 = 6 x 6 = 36

Case 3: Exactly 3 senior partners

In this case, we need to pick 3 senior partners from 4 senior partners and no junior partners from 6 junior partners. The number of ways this can be done is 4C3 x 6C0.

4C3 x 6C0 = (4x3x2)/3! x 1 = 4 x 1 = 4

Thus, the total number of ways to form a group in which there is at least 1 senior partner = 60 + 36 + 4 = 100.

Alternate Solution:

It must be true that:

The total number of ways to form a group of 3 partners = (The number of ways in which the group would have at least 1 senior partner) + (The number of ways in which the group would have no senior partners).

Therefore:

The number of ways in which the group would have at least 1 senior partner = (The total number of ways to form a group of 3 partners) - (The number of ways in which the group would have no senior partners).

If the group of 3 has all junior partners, and there are 6 junior partners total, then the group of all junior partners can be made in 6C3 ways.

6C3 = (6 x 5 x 4)/3! = 5 x 4 = 20

The total number of groups of 3 that can be formed from 10 partners is 10C3.

10C3 = (10 x 9 x 8)/3! = 5 x 3 x 8 = 120

Thus, the number of ways to form a group of 3 in which there is at least 1 senior partner = 120 - 20 = 100 ways.

Answer: B

Your explanation says that there can be all 3 senior members also as mentioned AT LEAST 1 senior so there can be 3 at most , FINE

But in the last line (2 groups are considered different if at least one group member is different), if we consider this , then it means we need atleast one member should be different in a group ,

So 4C1 * 6C2 (1 senior 2 junior) + 4C2 * 6C1 (2 senior and 1 junior) + 4C3 (Why this 4C3? As we need atleast one different member but in this 4C3 we will have all seniors and no junior and this will contradict the last line)

Please help?

2 groups are considered different if at least one group member is different does NOT mean that the group must have members from both seniors and juniors. It means that (ABC) and (ABD) considered different groups because at least one group member is different, while (ABC) and (ACB) are NOT considered different because all members of these groups are the same even the order is different. This part (2 groups are considered different if at least one group member is different) is added there to hint that the order of the group is not important and only the members of the group are.

Hope it's clear.

I don't understand why this is not valid:
a) Pick one senior partner from 4
b) Pick 2 x further partners from 9 remaining senior & junior partners

4 (senior partner) * 9 (other partner) * 8 (other partner) = 288

Order of selection does not matter, so divide by # of ways to re-arrange 3 partners i.e. 3! = 6

288 / 6 = 48

But this is the wrong answer. Why?

Hi Manychips,

In your calculation, you assume that the first person chosen MUST be a Senior Partner, but then you divide the entire calculation by 3! (which is something that you can only do if the first person can be ANY partner - Senior or Junior). If you want to approach the question in this way, then you would have to do the following:

Total number of PERMUTATIONS (regardless of whether the member is Senior or Junior) = (10)(9)(8) = 720
Total number of UNIQUE groups of three = 720/3! = 120

Total Permutations of three that are ONLY Junior Members = (6)(5)(4) = 120
Total number of UNIQUE groups of Junior Members = 120/3! = 20

Total groups with AT LEAST 1 Senior Member = (Total of All Groups) - (Total of JUST Junior Members) = 120 - 20 = 100

GMAT assassins aren't born, they're made,
Rich

Please tell me what I have done wrong
First , we choose a Senior member :
C(4,1) =4
then we choose 2 members from the rest (9)
C(9,2) =36
so the number of methods is 4*36 = 144

Wrong but why ?

Hi foryearss,

Unfortunately, your math includes some 'duplicate entries.'

For example, let's call the 4 senior partners A, B, C and D and the 6 junior partners 1, 2, 3, 4, 5 and 6.

In your calculation, you state that the first person selected MUST be one of those 4 seniors (A/B/C/D) and the remaining two people can be any two of the remaining 9...

The group "A/B/1" and "B/A/1" are the SAME group, but your calculation counts THAT group TWICE (depending on whether A or B was chosen first). In a Combination question, you can't allow duplicate entries.

GMAT assassins aren't born, they're made,
Rich

Bunuel VeritasKarishma I got a doubt regarding an alternate approach. Would really appreciate your help

Let Senior Partner = S, Junior Partner = J

Approach: Choose 1 S and then consider the other cases.

Ways of choosing 1 S: 4C1 = 4
Then we have 3 S left

Now the possible cases of picking 2 people from 3S and 6J are: 1) 1S + 1J 2) 0S + 2J 3) 2S + 0J
Total # for those 3 cases = 3C1 * 6C1 + 6C2 + 3C2 = 3*6 + 15 + 3 = 36

Total cases = 4 * 36 = 144

What am I doing wrong since the answer doesn't match the OA? Thanks!