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choose a five digit lock code in which the first e last digit must be ODD and aren’t allowed repetition:

the result is 8*7*6*5*4=6720 and until now is okay..

But is it possible to calculate this result in this way:?

1)calculate all the possible arrangements with 5 digits= 10!/5! 2) calculate all the numbers that have in the first and in the last place an EVEN number.

then subtract 1)-2) and you should have the result above.. why this reasoning is flawed?

choose a five digit lock code in which the first e last digit must be ODD and aren’t allowed repetition:

the result is 8*7*6*5*4=6720 and until now is okay..

But is it possible to calculate this result in this way:?

1)calculate all the possible arrangements with 5 digits= 10!/5! 2) calculate all the numbers that have in the first and in the last place an EVEN number.

then subtract 1)-2) and you should have the result above.. why this reasoning is flawed?

Because it gives permission to repetition of numbers.

choose a five digit lock code in which the first e last digit must be ODD and aren’t allowed repetition:

the result is 8*7*6*5*4=6720 and until now is okay..

But is it possible to calculate this result in this way:?

1)calculate all the possible arrangements with 5 digits= 10!/5! 2) calculate all the numbers that have in the first and in the last place an EVEN number.

then subtract 1)-2) and you should have the result above.. why this reasoning is flawed?

Because it gives permission to repetition of numbers.

sorry but where? 10!/5! is like ABCDEXXXXX where A,B,C,D,E are 5 different numbers choosen in a gropu of ten..

choose a five digit lock code in which the first e last digit must be ODD and aren’t allowed repetition:

the result is 8*7*6*5*4=6720 and until now is okay..

But is it possible to calculate this result in this way:?

1)calculate all the possible arrangements with 5 digits= 10!/5! 2) calculate all the numbers that have in the first and in the last place an EVEN number.

then subtract 1)-2) and you should have the result above.. why this reasoning is flawed?

Because it gives permission to repetition of numbers.

sorry but where? 10!/5! is like ABCDEXXXXX where A,B,C,D,E are 5 different numbers choosen in a gropu of ten..

But the formula you are giving is not the formula of combination it is the formula of permutation.

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