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4 professors and 6 students are being considered for

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Re: 4 professors and 6 students are being considered for [#permalink] New post 08 Sep 2013, 22:53
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tyagigar wrote:
why i can not do like this:

formation has to have at-least 1 professor :
we chose no of ways i can select 1 professor : 4c1= 4 ways
no we have to select 2 more people and since our requirement of atleast 1 professor is already met can we not select the remaining 2 people out of 9 in 9c2 ways i.e (9*8*7!)/(2!*7!)=36 ways
so total no of ways=4*36

i know i am doing something wrong , can i get some help please


The number you get will have duplications.

Consider the group of three professors {ABC}
Say you select professor A with 4C1, next you can get professors B and C with 9C2: {A}{BC}. Now, with the same method you can get the following group {B}{AC}, which is basically the same group.

Does this make sense?
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Re: 4 professors and 6 students are being considered for [#permalink] New post 09 Sep 2013, 06:35
Bunuel wrote:
tyagigar wrote:
why i can not do like this:

formation has to have at-least 1 professor :
we chose no of ways i can select 1 professor : 4c1= 4 ways
no we have to select 2 more people and since our requirement of atleast 1 professor is already met can we not select the remaining 2 people out of 9 in 9c2 ways i.e (9*8*7!)/(2!*7!)=36 ways
so total no of ways=4*36

i know i am doing something wrong , can i get some help please


The number you get will have duplications.

Consider the group of three professors {ABC}
Say you select professor A with 4C1, next you can get professors B and C with 9C2: {A}{BC}. Now, with the same method you can get the following group {B}{AC}, which is basically the same group.

Does this make sense?


thanks a lot, this makes perfect sense
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Re: 4 professors and 6 students are being considered for [#permalink] New post 22 Apr 2014, 02:47
Can any one tell where the below approach went wrong?

Selecting one professor from 4 * select any two from the remaining nine members
4c1 * 9c2 = 4 * 9*8/2*1 = 144
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Re: 4 professors and 6 students are being considered for [#permalink] New post 22 Apr 2014, 02:53
Expert's post
saiyanVegeta wrote:
Can any one tell where the below approach went wrong?

Selecting one professor from 4 * select any two from the remaining nine members
4c1 * 9c2 = 4 * 9*8/2*1 = 144


Check here: 4-professors-and-6-students-are-being-considered-for-85865-20.html#p1265096

Hope it helps.
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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: 4 professors and 6 students are being considered for [#permalink] New post 25 Feb 2015, 10:28
In this problem, may I ask why we divided by the part in red?

10!/(7!3!) - 6!/(3!3!) = 100

I understand the idea, and the first part is perfectly clear. I understand subtracting the possibility of no teachers, but what does 3!3! mean?
Now, I understand why it is not 6!/3!, but in this case 3!3! seems to mean 3 students selected and 3 not selected. Or not...?
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Re: 4 professors and 6 students are being considered for [#permalink] New post 25 Feb 2015, 16:11
Expert's post
Hi pacifist85,

The approach that you are referencing is meant to do the entire calculation in one gigantic "move", which I'm not a big fan of. It's simple enough to break the calculation into 3 "pieces" and then add together the results.

However, since you asked though, this calculation is based on two ideas:

1) You have to choose 3 people from a group of 10 (that's the first part of the calculation)
2) Since the group has to include at least 1 professor, you CAN'T have a group that is made up of 3 students.

Since there are 6 total students, there are 6!/3!3! different groups of 3 students that could be formed. We have to REMOVE those options from the prior part of the calculation. This gives us:

(All possible groups of 3) - (All groups of 3 that are JUST students) = (groups of 3 with at least 1 professor).

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Re: 4 professors and 6 students are being considered for [#permalink] New post 27 Feb 2015, 11:30
all possibitlity: C(10 3) = 120
possibility for no professor: C(6 3) = 20
possibility for at least one professor = 120 - 20 1= 100
Re: 4 professors and 6 students are being considered for   [#permalink] 27 Feb 2015, 11:30

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