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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
Expert Post
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Joined: 02 Sep 2009
Posts: 23422
Followers: 3619

Kudos [?]: 28987 [0], given: 2874

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.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

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.


What are GMAT Club Tests?
25 extra-hard Quant Tests

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Re: 4 professors and 6 students are being considered for   [#permalink] 22 Apr 2014, 02:53
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