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Consider a situation where we have telephone numbers consisting of 7 digits. How many telephone numbers are there which do NOT include any other digits but 2, 3, 5 and 7?

The question does not mention anything about repetition.

To solve this, I tried $(total\ possible\ telephone\ numbers) - (numbers\ which\ contain\ 1,\ 4,\ 6,\ 8,\ 9,\ and\ 0)$ but that just complexifies it (i think).

Another way to solve this might be to make cases.

There can be many cases:

  1. 3 digits repeat and 1 doesn't. For eg: $2\ 2\ 3\ 3\ 5\ 5\ 7$

  2. 1 digit comes thrice, 1 digit twice, and 2 digits only once. For eg: $2\ 2\ 2\ 3\ 3\ 5\ 7$

I can think of many more cases but there are just so many and I don't know how to proceed. Can someone provide a hint/solution and also tell me how to tackle such combinatorics problems?

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    $\begingroup$ There are four distinct digits, and each can be in any of the places, so $4^7$ $\endgroup$
    – abiessu
    Feb 4 at 8:04

1 Answer 1

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We have 7 slots for numbers

$$\,\,\_ \,\,\_ \,\,\_ \,\,\_ \,\,\_ \,\,\_ \,\,\_ $$

For each slot, we have $4$ different independent choices for numbers.

Hence, the total number of ways to choose $7$ numbers independently with $4$ choices for each number is simply given by multiplying as follows

$$4\cdot 4\cdot 4\cdot 4\cdot 4\cdot 4\cdot 4=4^7=16384 $$

which is as @abiessu has mentioned in the comments.

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