Its a permutation or comibination related question?

A nurse is preparing program to meet expectant mothers,she has 4 topics to cover.
a)How many different program is it possible for her to prepare?
b)Suppose at the last minute she finds that she has time for only three topics .How many different programs is it possible for her to prepare if she consider all four of equal importance?

a)it's permutation 4 x 3 x 2 x 1=24
b)it's combination 4!/ 3! x 1!=4

Dear Caramel thanx for solving my question. Report It

No problem.The same day I was studying for my maths test the same problems ann I am glad I helped :) Report It

order is important ..
so 4*3*2*1 = 24

if she only has time for three
4*3*2 = 24

She could choose any of the 4 to start off, then any of the remaining 3, then either of the remaining 2, then the last one for 4x3x2x1=24 ways.

If only 3 are used, it's still the same because only the "times 1" at the end would be omitted.

permutation

a) This is a permutation question, the order of the topics is important...4 P 4 = 4!/(4-4)! = 4!/0! = 4!/1 = 4! = 4*3*2*1 = 24. There are 24 different programs that she could create.
b) If the nurse considers the order of topics important, then this is a permutations problem - 4 P 3 = 4!/(4-3)! = 4!/1! = 4!/1 = 4! = 4*3*2*1 = 24 and she can still create 24 different programs. However, if the order of topics is not important and she simply wants to know how many ways she can chose 3 topics out of 4, it's a combinations problem 4 C 3 = 4! / ((4-3)! * 3! ) = 4! / (1! * 3!) = 4*3*2*1 / (1*3*2*1) = 4.

a) 4! = 24
b) 4C3 = 4

This is an interesting question. It sounds like you are considering any different ordering of topics to be a different program. Therefore order is important for the question at hand. Therefore you are concerned with a Permutation problem. The easiest way to think about it though is that she has 4 different choices for the first topic, 3 different choices for the second, 2 different choices for the third topic and finally just one choice for the last.

Therefore, there are 4*3*2*1=24 different possible programs when order is important.

b) The second question could be taken two different ways. If we again assume that the order of the topics is important (and would constitute a different program) then this is again a permutation question. There would be nPr=4 P 3 =24 different programs as well. Or you could think of it as: 4 different choices for this first topic, 3 different choices for the second topic, 2 different choices for the third topic.

Therefore overall there is 4*3*2=24 different programs available.

If the nurse does not think the order of the topics is important that would change this problem to a combination question. In that case there would be nCR r= 4 C 3 = 4 different programs available.