Helpful question: In how many ways can we make or arrange five digits so that there is no repetition?
We know that there are 10 possible digits to choose from for the first digit, 0 - 10. However, we can't have the first digit be a 0, else we get a four-digit number. So, we're down to 9. For each of the 9 digits, there are 9 digits to choose from for the second digit to avoid repetition and so on.
This gives $9 \cdot 9 \cdot 8 \cdot 7 \cdot 6 = 27,216$ ways.
Actual question: How many five-digit numbers have at least one digit that occurs more than once?
We also know that there are $9 \cdot 10 \cdot 10 \cdot 10 \cdot 10 = 90,000$ possible five-digit numbers. At this point, it should be clear why the first digit can only be selected in 9 ways.
In order to get the number of five-digit numbers that have at least one digit repeated, we simply subtract the number of possible numbers without repeated digits from the total number of five-digit numbers, which gives us $90,000 - 27,216 = 62,784$
My answer is probably redundant at this point. Nevertheless, I'll just leave it here. :]
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Question
The number of five digit telephone numbers having at least one of their digits repeated is
-
90000
-
100000
-
30240
-
69760
easy
Solution
Correct option is
69760
The number of five digit telephone numbers which can formed using the digits 0, 1, 2, … 9 is 105 [see example 19]. The number of five digit telephone numbers which have none of their digits repeated is 10P5 = 30240. Thus, the required number of telephone numbers is 105 – 30240 = 69760.
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