Question

Two red counters, three green counters and 4 blue counters are placed in a row in random order.
The probability that no two blue counters are adjacent is
A. $\dfrac{9}{{42}}$
B. $\dfrac{5}{2}$
C. $\dfrac{5}{{42}}$
D.None

Answer 1

Hint : In this question, we are given data about the counters and one condition for arranging the counters. We have to find the probability of that condition being followed. Probability is defined as the chances of occurring in an event. It is equal to the ratio of the number of possible outcomes and the total number of outcomes.

Complete step by step solution:
We are given counters of three colours and the counters. There are two red counters, three green counters and 4 blue counters, and they are placed in a row.
Total number of counters = 9
When the blue counters are arranged without any condition, the total ways of arranging them is $^9{C_4}$
In this question, we are given a condition that no two blue counters are adjacent, that is, two blue counters cannot be next to each other. So we can show the arrangement as –
$0 \times 0 \times 0 \times 0 \times 0 \times 0$
Where $0$ represents a blue counter and $\times$ represents a red or green counter.
So, we see that there are 6 possible positions of the blue counter.
Thus, ways of arranging the counters by the given condition is $^6{C_4}$
Now,
$Probability = \dfrac{{number\,of\,possible\,ways}}{{total\,ways}} \\\ \Rightarrow Probability = \dfrac{{^6{C_4}}}{{^9{C_4}}} \;$
We know that $^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$
So, we get –
$\Rightarrow Probability = \dfrac{{\dfrac{{6!}}{{2!4!}}}}{{\dfrac{{9!}}{{5!4!}}}} \\\ \Rightarrow Probability = \dfrac{{6!}}{{2!4!}} \times \dfrac{{5!4!}}{{9!}} \\\ \Rightarrow Probability = \dfrac{{6!}}{{2 \times 1{\kern 1pt} }} \times \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{9 \times 8 \times 7 \times 6!}} \\\ \Rightarrow Probability = \dfrac{{5 \times 4 \times 3}}{{9 \times 8 \times 7}} \\\ \Rightarrow Probability = \dfrac{5}{{42}} \;$
Hence, option (C) is the correct answer.
So, the correct answer is “Option C”.

Note : For finding the probability of any event, we have to first find out the number of possible outcomes and total numbers of outcomes as we found in this question. $^n{C_r}$ represents the combination of “r” terms out of total “n” terms, that is, it tells us the number of ways in which “r” out of “n” terms can be arranged.