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# How many ways to get a 6-digit telephone number which has at least one of their digits repeated ?

Last updated date: 13th Jul 2024
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Hint: Let us find the number of ways in which no digit is repeated and then subtract that by the total number of ways of forming a 6-digit number.

As we know that we are asked to find a number of ways when at least one digit is repeated. And this can also be written as the number sum of ways when exactly one digit is repeated, when exactly two digits are repeated, when exactly three digits are repeated, when exactly four digits are repeated, when exactly five digits are repeated and when all six digits are repeated.
So, this means we had to exclude ways in which no digit is repeated from the total number of ways.
So, there are a total 10 possible digits for each of the six places and that were {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.
So, total number of ways of forming 6-digit telephone number without any restriction will be 10*10*10*10*10*10 = 1000000.
And if all digits are different,
Then the number of possible digits for ${1^{st}}$ place will be 10.
Number of possible digits for ${2^{nd}}$ place will be 9.
Number of possible digits for ${3^{rd}}$ place will be 8.
Number of possible digits for ${4^{th}}$ place will be 7.
Number of possible digits for ${5^{th}}$ place will be 6.
Number of possible digits for ${6^{th}}$ place will be 5.
So, total number of ways for forming a 6-digit number in which all digits are different will be 10*9*8*7*6*5 = 151200.
So, the total number of 6-digits telephone numbers possible in which at least one digit is repeated will be 1000000 – 151200 = 848800.
Hence, the correct answer will be 848800.

Note: Whenever we come up with this type of problem where we are given that at least 1 digit is repeated then we should find the number of ways in which all the digits are different and then subtract that by the total number of possible ways without any restriction. This will be the easiest and efficient way to find the solution of the problem.