any one help me...
HOw many committees can be formed with at most 5 members?
Would the 5-man committe include all the persons we could choose from?
yes, could you tell me how many? and how did you get the answer?
Nope. Instead, try to write down how you should approach the problem, identify your particlur difficulties, THEN I'll come to your rescue.
That holds for your other thread as well.
oh come on... i dont know how... pls help me Mr. Arildno...
Either a guy is on the committee, or he is not.
That holds for each guy!
Try to think what this implies.
could it be 5 committees?
Try to apply the multiplication principle instead!
It is easiest first to think of "the committe with 0 members" as a committee itself when setting up the calculation; when you have done that, just subtract one for your answer.
This won't work, diceyfume.
You are obviously refusing to utilize your intellect, and are only interested in being spoonfed "answers".
I strongly advise you to quit maths, because your attitude makes you incompetent in it.
i guess i know the answer... 5*4*3*2*1
120 is the answer?
Not quite, but a definite improvement!
Sorry about my prior annoyance at you; I hope you will continue to visit PF in the future.
Try to think about this problem in another way:
Call the 5 persons A,B,C,D,E.
We are to form a committee out of these, where the committe can consist of 1 to 5 persons.
Either, A is included in the committe, or not (2 classes of committees thereby delineated)
Similarly with the four others.
Thus, one might think the answer should be:
2*2*2*2*2=32 different comittees, but one of those committees consists of no members at all!
Thus, the correct answer must be 32-1=31 committees in total.
Now, you might wonder: Why is your approach wrong?
Suppose you have the following case:
Exactly 5 places exist on the comittee, and each place has a unique function
(say, chairman, vice-chairman, secretary, accountant and public relations guy).
In this case, say we elect in this order:
Then, there are 5 different possibilities for chairman, once he has been chosen, 4 different choices for vice-chairman (yielding 5*4=20 different two-man groups) and so on.
In this case, your answer would be the correct one!
But, that is not at all what is presupposed in this exercise!
First off, committees can be of VARIABLE size.
Secondly, each committee member has basically the same function as any other.
Thus, this is a totally different scenario from the case in which your calculations would be correct.
Hope that helps.
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