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

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Re: A firm has 4 senior partners and 6 junior partners. How many different  [#permalink]

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22 May 2018, 10:12
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

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Re: A firm has 4 senior partners and 6 junior partners. How many different  [#permalink]

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28 May 2018, 20:22
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 ?
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Re: A firm has 4 senior partners and 6 junior partners. How many different  [#permalink]

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28 May 2018, 21:11
1

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.

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Re: A firm has 4 senior partners and 6 junior partners. How many different  [#permalink]

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20 Aug 2019, 04:07
gdk800 wrote:
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

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!
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Re: A firm has 4 senior partners and 6 junior partners. How many different  [#permalink]

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20 Aug 2019, 05:31
gdk800 wrote:
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

Given: A firm has 4 senior partners and 6 junior partners.

Asked: How many different groups of 3 partners can be formed in which at least one member of the group is a senior partner?

Number of groups in which at least one member is a senior partner = Total number of groups - Number of groups in which no member is a senior partner

Total number of groups $$= ^{10}C_3 = 120$$
Number of groups in which no member is a senior partner $$= ^6C_3 = 20$$

Number of groups in which at least one member is a senior partner = Total number of groups - Number of groups in which no member is a senior partner
= 120 - 20 = 100

IMO B
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Re: A firm has 4 senior partners and 6 junior partners. How many different  [#permalink]

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21 Aug 2019, 00:13
2
dabaobao wrote:
gdk800 wrote:
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

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!

You are double counting cases. Whenever you make partial selections from a group for another group, you double count. Here, you first selected an S for your team and then selected 1 S or 2S or 0S again from the group of S for your team.

So say the 4 S are Sa, Sb, Sc and Sd. You select Sa.
Now in your first case, you have to select another S. Say you select Sc. Then you select Ja from juniors.

Now think of another case. First you select Sc.
Now in your first case, you select Sa. Then you select Ja from juniors.

Both these teams are the same but you counted them as 2. That's your error.
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Re: A firm has 4 senior partners and 6 junior partners. How many different  [#permalink]

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21 Aug 2019, 01:13
gdk800 wrote:
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

3 people can be selected out of 10people in 10C3 ways =120 ways.

Out of these 120 ways there will be some group where there is no senior.

No of groups where there is no senior = 6C3= 20

So groups where there will be at least one senior = 120 -20
=100

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Re: A firm has 4 senior partners and 6 junior partners. How many different  [#permalink]

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23 Aug 2019, 07:09
gdk800 wrote:
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

we get 3 possiblities ; 4c1*6c2+ 4c2*6c1+ 4c3 ; 100
IMO B
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Re: A firm has 4 senior partners and 6 junior partners. How many different  [#permalink]

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26 Sep 2019, 05:51
statement in the brackets is very confusing. can anyone explain ?
as per me answer should be 96. we cannot consider the case third because all the member will be seniors and thus 4 groups wont be considered different.
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Re: A firm has 4 senior partners and 6 junior partners. How many different  [#permalink]

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26 Sep 2019, 11:58
1
Hi ashusharma2965,

The wording in parentheses is there to help us eliminate 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.
The group "A/B/1" and the group "B/A/1" are the SAME group, since they have the same 3 members - thus, we are NOT supposed to count that group more than once.

Based on the wording in the prompt, there are 3 types of groups that "fit" what we are looking for.

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

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Re: A firm has 4 senior partners and 6 junior partners. How many different  [#permalink]

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17 Oct 2019, 07:24
Bunuel wrote:
gdk800 wrote:
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

Total # of different groups of 3 out of 10 people: $$C^3_{10}=120$$;
# of groups with only junior partners (so with zero senior memeber): $$C^3_6=20$$;

So the # of groups with at least one senior partner is {all} - {none} = {at least one} = 120 - 20 = 100.

Hi Bunuel,

I picked 288 and got it wrong. Here is my though process, if you can point out what I did wrong that would be great.

for the three spot, the first spot is one of four senior partners, since I fullfilled the "at least one senior partner" requirement at spot one, the rest two spots can mix and match senior and junior partners. The second
as one out of nine options and the third spot has eight options. So I have 4*9*8=288. Did I miss anything?

Thanks!
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Re: A firm has 4 senior partners and 6 junior partners. How many different  [#permalink]

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17 Oct 2019, 09:11
gdk800 wrote:
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. (Two groups are considered different if at least one group member is different)

A. 48
B. 100
C. 120
D. 288
E. 600

Given: A firm has 4 senior partners and 6 junior partners.

Asked: 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)

1S2J = 4*6C2 = 4*15 = 60
2S1J = 4C2*6 = 36
3S = 4C3 = 4

Total ways = 60 + 36 + 4 = 100

IMO B

Alternatively

No of ways = total ways - no senior partner
No of ways = 10C3 - 6C3 = 10*9*8/3*2*1 - 6*5*4/3*2*1 = 120 - 20 = 100

IMO B
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Re: A firm has 4 senior partners and 6 junior partners. How many different  [#permalink]

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13 Nov 2019, 02:46
Bunuel wrote:
gdk800 wrote:
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

Total # of different groups of 3 out of 10 people: $$C^3_{10}=120$$;
# of groups with only junior partners (so with zero senior memeber): $$C^3_6=20$$;

So the # of groups with at least one senior partner is {all} - {none} = {at least one} = 120 - 20 = 100.

Shouldn't we go for permutations rather than combination here ?
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Re: A firm has 4 senior partners and 6 junior partners. How many different  [#permalink]

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13 Nov 2019, 19:12
1
Hi dasoisheretorule,

Many questions that can be approached as a Permutation or Combination can ALSO be approached in the other way (it's just a matter of properly organizing your calculations). This prompt specifically refers to the number of "different GROUPS" (meaning that the 'order' of the members does NOT matter) - and that's a common clue that we're dealing with a Combination Formula question.

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Re: A firm has 4 senior partners and 6 junior partners. How many different   [#permalink] 13 Nov 2019, 19:12

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