Question

How many 3-digit numbers can be formed from the digi's 2, 3, 5, 6, 7 and 9? Each digit used only once
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Answer 1

Answer:

120

Step-by-step explanation:

we have 6 numbers

at first 6 number after that 5 because number will not repeat than 4

6×5 ×4=120

Answer 2

Answer:

the answer is 120.

Step-by-step explanation:

Let there are three places to fill...

H T U   (H stands for Hundred place, T stands for tens place and U stands for unit place.)

For place H, there are 6 digits available out of which any one digit will be selected.So, there will be total 6 different ways for putting a digit at H place.

Similarly, For place T, there are 5 digits available (because one digit is already selected for H place) , out of which any one digit will be selected. So, there will be total 5 different ways for putting a digit at T place.

And,  For place U, there are 4 digits available (because two digits are already selected for H & T places) , out of which any one digit will be selected. So, there will be total 4 different ways for putting a digit at T place.

So, there will be total 6*5*4 = 120 3-digits numbers can be formed without repetition.

This Problem can be solved easily by Permutation as:

⁶P₃ = 6!/(6-3)! = 6*5*4*3!/3! = 6*5*4 = 120

Hope You understood the concept !!!