Question

A child has 12 blocks, of which 6 are black, 4 are red, 1 is white, and 1 is yellow. (a) If the child puts the blocks in a line, how many different arrangements are possible? (b) If one of the arrangements in part (a) is randomly selected, what is the probability that no two black blocks are next to each other.

Answer 1

a) The number of different arrangements possible here is computed as:

Note that denominator as 6! for 6 identical black blocks and 4 for 4 identical red blocks

b) Here first we compute the number of ways in which blocks could be placed such that no black is together. This is computed as:

First arrange 4 red, 1 white and 1 yellow in 6! / 4! = 30 ways

Then after this arrangement, there are 7 places where the 6 black blocks could be placed so that none of them are adjascent to each other.

Therefore total number of ways of such arrangements here are given as:

= 30*(7c6) = 210 ways

The required probability thus is computed as:

Therefore 0.0076 is the required probability here.