Question

The number of 4 digit numbers that can be formed from the digits 0, 1, 2, 3, 4, 5, 6, 7 so that each number contains digit 1 is
A. 1225
B. 1252
C. 630
D. 480

Answer 1

Hint: Use the formula for the total number of ways to arrange ‘n’ different objects in ‘r’ places and calculate the number of ways to arrange the 8 digits in four places with the digit 1 present each time. Then exclude the possibilities of digit 0 being at the beginning.
Formula used:
 $ {}^{n}{{P}_{r}}=\dfrac{n!}{(n-r)!} $ .

Complete step-by-step answer:
We have 8 digits 0, 1, 2, 3, 4, 5, 6, 7. We are supposed to find the number of ways to arrange them such that they form a four digit number with digit 1 in it compulsorily.
This means that in every arrangement the digit 1 will be there. Therefore, let us fix at the digit 1. There are four ways to place the digit 1 since it is a 4 digit number.
Suppose the digit 1 is at the beginning of the number (i.e. at the ten-thousandth place).
1 __ __ __
Now, we have to arrange the remaining 7 digits in three places.
The total number of ways to arrange ‘n’ different objects in ‘r’ places is equal to $ {}^{n}{{P}_{r}}=\dfrac{n!}{(n-r)!} $ .
The total number of ways to arrange ‘n’ different objects in ‘r’ places is equal to $ \Rightarrow {}^{n}{{P}_{r}}=\dfrac{n!}{(n-r)!} $ .
In this case, $ n=7 $ and $ r=3 $ .
Therefore, the total number of ways to arrange the 7 digits in 3 places is equal to $ \Rightarrow {}^{7}{{P}_{3}}=\dfrac{7!}{(7-3)!}=\dfrac{7!}{4!}=7\times 6\times 5=210 $ .
Similarly, the digits 1 can be at other three places each time, which will give us more 210 ways three times.
Hence, the total number of ways to arrange the given digits with the digit 1 present every time is $ 4\times 210=840 $ .
We want a four digits number. However, there are possibilities where the digit 0 will come at the beginning and it will become a three digit number. Hence, we have to exclude such possibilities.
The number of numbers with digit 0 in the beginning will be equal to $ \Rightarrow {}^{7}{{P}_{3}}=\dfrac{7!}{(7-3)!}=\dfrac{7!}{4!}=7\times 6\times 5=210 $ .
Therefore, the total number of 4 digit numbers that can be formed from the digits 0, 1, 2, 3, 4, 5, 6, 7 so that each number contains digit 1 is $ 840-210=630 $ .
So, the correct answer is “Option C”.

Note: Whenever a question is of this type, where we have to find the number of ways to arrange some objects under a given condition, the best way is to calculate the total number of arrangements possible and then subtract the possibilities that do not satisfy the condition.