Question

Numbers 1, 2, 3, 4, 5, 6, 7, 8 are arranged in random order. The probability that the digits 1, 2, 3, 4 appears as neighbours in that order, is
(a) $\dfrac{1}{2}$
(b) $\dfrac{1}{128}$
(c) $\dfrac{1}{256}$
(d) $\dfrac{1}{336}$

Answer 1

Hint: We start solving the problem by finding the total arrangements that we can made using the digits 1, 2, 3, 4, 5, 6, 7, 8 by using the fact that the total number of ways of arranging ‘n’ objects in ‘n’ places is $n!$. We then assume the arrangement $1234$ as a single digit and find the total number of arrangements that we can do in using the arrangement $1234$ and other digits. We then recall the formula of probability as $\text{probability=}\dfrac{\text{favorable arrangements}}{\text{total number of arrangements}}$ and substitute the values in it to get the required answer.

Complete step by step answer:
According to the problem, we are given that the numbers 1, 2, 3, 4, 5, 6, 7, 8 are arranged in random order and we need to find the probability that the digits 1, 2, 3, 4 appear as neighbours in that order.
Let us find the total number of arrangements that we can make with the digits 1, 2, 3, 4, 5, 6, 7, 8.
We know that the total number of ways of arranging ‘n’ objects in ‘n’ places is $n!$.
So, we can arrange the digits 1, 2, 3, 4, 5, 6, 7, 8 in $8!$ ways ---(1).
Now, let us find the total no. of favorable arrangements we can make with the digits 1, 2, 3, 4 appearing as neighbours in that order.
Let us assume that $1234$ as a single digit ‘x’ as we need in that order and does not need any arrangements in between them.
So, we have to arrange the digits x, 5, 6, 7, 8 which will be our favorable arrangements.
The total no. of favorable arrangements is $5!$ ways ---(2).
We know that probability of an event is defined as $\text{probability=}\dfrac{\text{favorable arrangements}}{\text{total number of arrangements}}$.
So, we get the required probability = $\dfrac{5!}{8!}$.
We know that $n!=n\times \left( n-1 \right)\times \left( n-2 \right)\times ......\times 2\times 1$.
$\Rightarrow $ Required probability = $\dfrac{5\times 4\times 3\times 2\times 1}{8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}$.
$\Rightarrow $ Required probability = $\dfrac{1}{8\times 7\times 6}$.
$\Rightarrow $ Required probability = $\dfrac{1}{336}$.

So, the correct answer is “Option d”.

Note: Here we are assuming that we are only finding the total 8-digit numbers formed. If we have also taken the 7-digit, 6-digit etc., the answer would have been different. Whenever we get this type of problem without mentioning the number of digits in the problem, we should take the maximum digit numbers that can be arranged unless it is mentioned in the problem. Similarly, we can expect problems to find the probability of a 6-digit number that can be formed with the digits 1, 2, 3, 4 appears as neighbours in that order.