Question

The number of five digit numbers that can be formed with 0, 1, 2, 3, 4, 5, 6, 7 is:
A. $ {}^{8}{{P}_{5}}-{}^{8}{{P}_{4}} $
B. $ {}^{8}{{P}_{5}}-{}^{7}{{P}_{3}} $
C. $ {}^{8}{{P}_{5}}-{}^{7}{{P}_{4}} $
D. $ {}^{8}{{P}_{5}}-{}^{6}{{P}_{3}} $

Answer 1

Hint: The given case is similar to the case of arranging ‘n’ different objects in ‘m’ places. Therefore, use the formula for permutation and calculate the number of five digit numbers that can be formed from the given 8 digits. Exclude the numbers beginning with 0.

Complete step-by-step answer:
There are 8 numbers (digits) given to us and that are 0, 1, 2, 3, 4, 5, 6, 7. We are asked to find the total number of five digit numbers that can be formed with the given 8 numbers. (let us assume that there is no repetition of any digit).
This can be thought as arranging 8 different objects in 5 different places. The number of ways for arranging (permutation) ‘n’ different objects in ‘m’ places is given as $ {}^{n}{{P}_{m}} $ .
In the given case, $ n=8 $ and $ m=5 $ .
Therefore, the number of ways to arrange the 8 digits in the 5 places is equal to $ {}^{8}{{P}_{5}} $ .
However, we can see that one of the eight digits is 0. If the digit ‘0’ comes at the beginning of the five digit number then it will not remain a five digit number and it will be a four digit number.
This means that we have to exclude the possibilities where ‘0’ comes at the beginning of the five digit number.
Let us fix the digit ‘0’ at the beginning and find the number of four digit numbers that can be formed.
0 _ _ _ _
Now, there are 7 digits and we have to arrange them in 4 places. The number of ways this can be done is equal to $ {}^{7}{{P}_{4}} $ .
Therefore, the total number of five digit numbers that can be formed from the 8 digits is $ {}^{8}{{P}_{5}}-{}^{7}{{P}_{4}} $ .
So, the correct answer is “Option C”.

Note: Permutation is the arrangement of different objects in the given number of places.
If we want to simplify the permutation further you can by using the formula for the permutation.
i.e. $ {}^{n}{{P}_{m}}=\dfrac{n!}{(n-m)!} $ .
Note that any permutation will result in a whole number only.