Question

Number of permutations of 1, 2, 3, 4, 5, 6, 7, 8, 9 taken all at a time such that the digit 1 appearing somewhere to the left of 2, 3 appears to the left of 4, 5 somewhere to the left of 6 is:
(a) 9.7!
(b) 8!
(c) 5!.4!
(d) 8!.4!

Answer 1

Hint: First, try to find the permutations of 1, 2, 3, 4, 5, 6, 7, 8, 9 taken all at a time. Then try to put on the other conditions given in the question. For instance, to apply the condition that digit 1 appears to the left of 2, multiply the total permutations by a factor of half, as there is an equal possibility of 1 appearing either to the left or to the right.

Complete step-by-step answer:
Starting with the solution, let us find the permutations of 1, 2, 3, 4, 5, 6, 7, 8, 9 taken all at a time.
We know the number of ways of arranging r objects from n different objects is given by $^{n}{{P}_{r}}$ .
Therefore, we can say that the number of ways of arranging nine different digits is $^{9}{{P}_{9}}$ .
Now try to find the number of arrangements which satisfy the conditions given in the question.
The first condition says digit 1 must appear somewhere to the left of 2.
So, in arranging the digits, there are two possibilities either 2 appears to the left of 1 else to the right, and both the options will have an equal number of possible arrangements.
Therefore, we can say that a factor of half will be multiplied by the number of total possible arrangements to satisfy the first condition.
Similarly, there are two possibilities of 3 appearing with respect to 4 either to the left else to the right. So, again a factor of half is to be multiplied to the total number arrangements to satisfy the second condition.
Now we need to multiply the total number of arrangements by another factor of half to satisfy the third condition, as an equal number of arrangements are possible with 5 being either to the left else to the right of 6.
Therefore, the permutations of the digits with all the necessary conditions satisfied is:
$^{9}{{P}_{9}}\times \dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}$
And according to the definition of permutation $^{n}{{P}_{r}}$ can be written as $\dfrac{n!}{\left( n-r \right)!}$
$\therefore {{\text{ }}^{9}}{{P}_{9}}\times \dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}=\dfrac{9!}{8\times (9-9)!}=\dfrac{9!}{8\times 0!}$
We know 0! = 1. Using which we can say:
The permutations of the digits with all the necessary conditions satisfied is:
$\dfrac{9!}{8\times 0!}=\dfrac{9\times 8 \times 7!}{8\times 1}=9\times 7!$
Therefore, the correct answer is option (a) $9\times 7!$ .

Note: In questions related to permutations and combinations, students generally face problems in deciding whether the question is based on selection or arrangement. Also, be careful about the repeated cases, as in problems related to permutations a general error is caused due to the cases which get repeated while using the formula. So, always cross-check your answer for any repeated cases if possible.