Question

How many positive integers less than 659 can be formed using 3, 5, 6 and 8 for it's digits, with each digit being used only once?​

Answer 1

Given:

numbers 3, 5, 6 and 8.

To Find:

How many positive integers less than 659 can be formed using 3, 5, 6 and 8 for it's digits, with each digit being used only once?​

Solution

according to question;

659  is a three digit number so possible positive integers less than 659 can be 1 digit, 2 digit, and 3 digit numbers(or integer) above 100 and less than 659.

Here, we have to form the numbers using 3, 5, 6 and 8.

repeatation is not allowed.

For 3 digit numbers:

There are three places for three digit numbers hundreds, tens and unit so

for the hundreds place except 8 all given number can be placed

so there are 3 possibilities  for hundreds place.

For the tens place there are 4 possibilities and

for ones place there are 4 possibilities too

⇒ total no. of possibilities is

$=3\times 4\times 4\\=48$  

For 2 digits :

there are 4 possibilities for tens place and 4 possibilities for units place

therefore, total no. of possibilities is

$=4\times4\\=16$

For 1 digit:

there is only 4 possibility

Therefore,  total no. of ways to arrange 3 , 5, 6 and 8 such that obtained number is less than 659 = 48 + 16 + 4 = 68