Question

Asked Nov 20, 2019

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how many six-digit numbers can be formed so that 2 are in the place of thousandth and 5 in the place of the hundredths?

would the formula go like P(6)_2 + P(6)_5 (permutation of six with two repeats and then plus five repeats) or 6!/2!*5! or none of these?

Step 1

First number can not be 0. So we can pick the fir...

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