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How many different four-digit even positive integers can be made using the digits 1, 2, 3, 4, 5 if no digit can be used more than once?

User KGhatak
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120 different four-digit even positive integers can be made using the digits 1, 2, 3, 4, 5 if no digit can be used more than once

Solution:

Given, digits are 1, 2, 3, 4, 5

We have to find the number of ways in which 4 digit number can be made form the given digits with out repetition.

Now, first we have to select 4 digits out of 5 digits to form a number.

As it is just combination, we can take in
^(5) \mathrm{C}_(4) \text { Ways. }


\mathrm{n} \mathrm{C}_{\mathrm{r}}=(n !)/((n-r) ! r !)


5 \mathrm{C}_(4)=(5 !)/((5-4) ! 4 !)=(5 * 4 !)/(1 ! 4 !)=5

So, we can select 4 digits in 5 ways.

And now, we can arrange them in 4! Factorial ways.

Then, total 5 x 4! Ways are available


5 * 4 !=5 !=5 * 4 * 3 * 2 * 1=120 \text { ways }

Hence, we can form 120 different numbers.

User Suhas Gavad
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