Question

How many 8 – digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 270 and no digit appears more than once?

Answer 1

Hint:Out of 8 – digit numbers we know the first 3 digits as 2, 7 and 0. Find the remaining ways to fill the 5 spaces. The digits 0 to 9 are 10 in numbers, out of which 3 are used and remaining digits are 7. Thus find how the 5 places can be filled by using 7 digits without repetition.

Complete step-by-step answer:
We need to construct an 8 – digit telephone number from the digits 0 to 9.
Thus counting all the digits, there are a total of 10 digits.
Now, all the telephone number’s start with 270 and there is no repetition of the digits. We can represent this as,
2 7 0 _ _ _ _ _
Out of the 8 – digit telephone number we know the first three digits as 2, 7, 0. Now we need to find the rest of the 5 – digit numbers.
We know that total digits = 10
Now we have used 3 of the digits. Now let us find the remaining digits.
Remaining digits = 10 – 3 = 7 digits.
Thus total possible 8 digit phone number can be formed by using,
2 7 0 1 2 3 4 5
The place marked 1 can be filled by 7 digits. Thus the place can be filled in 7 ways. Now place 2 can be filled in 6 ways with the 6 digits, as the digit can’t be repeated.
Now the place 3 can be filled in 5 ways with the remaining 5 digits and the place 5 can be filled in 3 digits in 3 ways.
Thus we can say in short that,
2 7 0 7 ways 6 ways 5 ways 4 ways 3 ways
$\therefore$ Total possible 8 digit phone number = $7\times 6\times 5\times 4\times 3=2520$ ways
Thus we can form 8 – digit telephone numbers using 0 to 9 digits in 2520 ways.

Note: You might think that you have 5 places and 7 numbers, thus can fill it in ${}^{7}{{C}_{5}}$ ways, which is wrong. For each place there are ways to fill. If the digits can be repeated, then the total possible ways will differ accordingly.