stne wrote:
Bunuel wrote:
ksharma12 wrote:
Consider the number of way a committee of 3 can be selected from 7 people- A,B,C,D,E,F,G if:
Order does not matter,
and C and E cannot be chosen together.
What is the justification?
C^3_7-C^1_5=30C^3_7 - # of ways we can choose any 3 out of 7 (without restriction).
C^1_5 - # of groups with C and E together (if C and E are in chosen group, then third member can be any out of 5 left, so total # of groups is
C^1_5).
just for discussion sake and to clear the concept I was wondering what would happen if the order mattered in the above scenario.
Eva ..Bunuel and others in the community request you to please contribute .
lets assume ABCDEFG is a 7 letter word so ABC is different from CAB meaning order matters
so if I wanted to form 3 letter words such that CE should not be together then , what would be the number of ways this could be done ?
I tried to a certain extent please verify if this is correct or not.
C^7_3 * 3! - C^1_1 *C^1_1*C^5_1*3!=180
C^7_3 = # of ways to select 3 out of 7
3! = # of ways to arrange them among themselves as order matters
C^1_1= choosing c
C^1_1 = choosing e
C^5_1= # of ways to select one from remaining 5 , after C and E have been chosen in the community.
3! = # of to arrange the 3 letter committee containing both C and E
Hope this is correct?
Yes, this is correct.
I just have my own preferences to count...so, for example, choose 3 out of 7 when order matters I write directly 7*6*5, meaning I directly take into account the order (why write the formula with the factorials for nCk and then multiply by k!, reduce...). First choice 7 options, second choice 6, third 5.
And I am not even writing the 1C1 factors. There is nothing wrong with it, but I know that C and E must be chosen, then I need just one extra person (letter
), so I can choose 1 out of 5, for which again I am not writing the 5C1, and then having C, E and * (somebody), I have to consider all the permutations of the three, so 3!...
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