Aptitude - Permutation and Combination - Discussion

The best way is,

6c3 = 20.

Thanks a lot for the explanation. It is easy now.

In the third explanation I didn't understand please explain it.

How many 5-digited numbers can be formed by using 0, 1, 2, 3, 4, 5, 6 which is divisible by 7 when repetition is allowed?

Please help me to get the answer.

Here, the answer is 20.

By using the fundamental principle of counting ie, if one activity can be done in 'm' num of different ways and the second activity can be done in 'n' number of different ways and these two activities can be done in MxN different ways.

For example, how mane 3 digit number can be formed By using the numbers 2, 3, 5, 8.without repeating.
Soln:-
Let the numbers are 2, 3, 5, 8 and u all know 3 digit number means it must have unit tens and hundred places.
So, now you just Make three column like H T U.
2 3 4 ie in unit place you can fill any of the 4 numbers means 4 different ways and remaining 2 place that is tens and hundred and in the tens place, you can fill in 3 different ways because repetition not allowed.
If once you used the digit you cannot use again. And next 100 places you can fill in 2 different ways.

Therefore total no of ways = 2 x 3 x 4.
= 24ways.
I Hope you understood.

Divisibility rule 3:

If the sum of the digit is multiple of 3 then that number is divisible by 3.

Ex: 123 is divisible by 3. 1 + 2 + 3 = 6 here 6 is multiple of 3 therefore 123 is divisible by 3.

Why can't we write 5c2? There are two places and 5 numbers.

Shortcut way to solve this problem :-6C3 => 6!/3!
= 6 * 5 * 4/3 * 2 * 1.
= 20.

Thanks @Rahul! good explanation.

There are 6 digits are given ,

As per question one(5) is fixed so 5! = 120.

Three digit number is going to be formed by 3! combination.

So, 120/6 = 20.