# HOW TO FIND HOW MANY 6 DIGIT NUMBERS CAN BE FORMED WITH THE GIVEN DIGITS

How to Find How Many 6 Digit Numbers Can Be Formed With the Given Digits :

Here we are going to see how to find how many 6 digit numbers can be formed with the given digits.

## How to Find How Many 6 Digit Numbers Can Be Formed With the Given Digits - Practice questions

Question 1 :

Each of the digits 1, 1, 2, 3, 3 and 4 is written on a separate card. The six cards are then laid out in a row to form a 6-digit number.

(i) How many distinct 6-digit numbers are there?

(ii) How many of these 6-digit numbers are even?

(iii) How many of these 6-digit numbers are divisible by 4?

Solution :

## (i) How many distinct 6-digit numbers are there?

In order to construct 6 digit number, let us write 6 dashes

____   ____   ____   ____   ____   ____

Number of ways to fill those 6 places  =  6!/2! 2!

=  (6 ⋅ 5 ⋅ 4  3  2 ⋅ 1)/(⋅ 1)(⋅ 1)

=  180 ways

## (ii) How many of these 6-digit numbers are even ?

Since the required number is even, we have two options to fill the unit places (2 or 4)

Case (i)

In case we use 2 in the unit place, we have 5 options (1, 1, 3, 3, 4)

number of ways  =  5!/2!2!  =  30

Case (ii)

In case we use 4 in the unit place, we have 5 options (1, 1, 3, 3, 2)

number of ways  =  5!/2!2!  =  30

Total number of ways in (i) and (ii)  =  30 + 30  =  60 ways

## (iii) How many of these 6-digit numbers are divisible by 4?

In the last two places, we have to use 12, 32 or 24.

Case (i)

Using 12 in the last two places.

If we use 12 to fill the last two places, then we will have 4 options(1, 3, 3, 4) to fill up the 4 places.

Number of ways  =  (4!/2!) ⋅ 1 ⋅ 1

=  12

Case (ii)

Using 32 in the last two places.

If we use 32 to fill the last two places, then we will have 4 options(1, 1, 3, 4) to fill up the 4 places.

Number of ways  =  (4!/2!) ⋅ 1 ⋅ 1

=  12

Case (iii)

Using 24 in the last two places.

If we use 24 to fill the last two places, then we will have 4 options(1, 1, 3, 3) to fill up the 4 places.

Number of ways  =  (4!/2!2!) ⋅ 1 ⋅ 1

=  6

Hence the required number of ways  =  12 + 12 + 6

=  30

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