shanmugamgsn wrote:

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 = 2.2.2.2.2 = 2^5

Hope this make sense to you.

-Shalabh

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