Question

How many 3-digit numbers can you form by using each of the digits 7,8,1 only

once?​

Answer 1

Answer:

178 ,781, 817, 871, 187, 881,771,111,. 888. 777 etc

Answer 2

Answer:

----- 6 -----

Explanation:

First,

We have got 3 digits respectively, 7,8,1

Then,

We can give a trial-and-error effort or simply using a Formulae. But first, let us use the trial-and-error method.

Trial-and-error Method

1. Now, let's put the digit 7 in the (_-_-_) first dash, which represents the number in the hundreds. (i.e. 7-_-_)
2. Now, let's put the other digits. And we find that the other 2 digits can only be placed in 2 different ways.... either (_-8-1) or, (_-1-8)
3. So, we find 2 numbers: (i) 781 and (ii) 718

Using the same method, only using different digits in the hundred's place, we might get:

• (i) 871, and
• (ii) 817         (Using '8' as the number in the hundreds)
• (i) 187, and
• (ii) 178         (Using '8' as the number in the hundreds)

So, the numbers are:

781, 718, 871, 817, 187, 178 = total ""6"" numbers (ANS)

Now, using the formulae

Formulae: If you have a number of digits, then count them...

Here, we have 3 digits.

Now, multiply all the numbers from 1 to that number of digit which is, in this case, '3'.

so, 1 × 2 × 3 = 6 (ANS)

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