I'm weak in Combinations & Probability
When i went through GMAT Math Book Combinations i have few queries in these examples
1. There are three marbles: 1 blue, 1 gray and 1 green. In how many ways is it possible to arrange marbles in a row?
Solution is : \(3!\)
2. There is a basket with 5 different fruits inside. A child is given the basket and he may pick out none, one or more of the fruits. In how many ways can this choice be made ?
Solution is : \(2^5\)
Both are combinations only.
Why this difference? Is this because arrangement and selection ?
Hi,Q 1. Arrangement-
When order is important, then we apply arrangement. The total possible arrangements are calculated via permutation formula- nPr.
Example- 3 marbles - 1 blue(B), gray(R) and 1 green(G) can be arranged like these ways. - BRG, BGR, GBR, GRB, RGB, RBG. So total 6 ways. Which is also given by 3! or 3P3 = 3.2.1 = 6 ways.
Alternatively, we can also understand this concept by following way.
# of ways to fill first place = 3 ( Any one of the 3 marbles can be picked up),
# of ways to fill second place = 2 ( Any one of the remaining 2 marbles can be picked up),
# of ways to fill third place = 1 ( Last marble will be picked up)
So total # of ways = 3.2.1 = 3!= 6 ways.Selection -
Selection is applicable when we do not lay any emphasis on arrangement. Only selection is important.
Example- In how many ways can one select 2 students out of 3 students, named Joe, David, & Luke?
Answer- The possible ways would be JD or DL or JL i.e. 3 ways only. JD or DJ is one and the same thing, hence we should not look into arrangement in this case.Q 2. This question is a sum of 5 combinations-
Selection of no fruit + Selection of 1 fruit + Selection of 2 fruits + Selection of 3 fruits + Selection of 4 fruits + Selection of 5 fruits
A very important word here is all 5 fruits are different.
a. Selection of first fruit = 2 ( It will be selected or not selected)
b. Selection of second fruit = 2 ( It will be selected or not selected)
c. Selection of third fruit = 2 ( It will be selected or not selected)
d. Selection of fourth fruit = 2 ( It will be selected or not selected)
e. Selection of fifth fruit = 2 ( It will be selected or not selected)
So total # of ways would be = a.b.c.d.e = 126.96.36.199.2 = 2^5
Hope this make sense to you.
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