Question

 How many 4 digits numbers divisible by 2 can be formed by digits 4,5,6,7,8 if the repetition of digits is not allowed ? *​

Answer 1

Step-by-step explanation:

Step-by-step approach

Let us understand the constraints first:

1. The leftmost digit cannot be zero

2. The last digit needs to be even

3. The sum of digits needs to be even

Now, let us understand firstly how to use the constraints in PnC (this would be more effective for the case of repetition not being allowed)

The first constraint is not something we must be beginning with, in other words we may or may not

The second constraint is something which limits the cases to half

The third one is basically a constraint that is focussed on the last choice we take. The last choice will be dependent on the sum of the previous three choices.

So let us select the digits in that order

Second constraint limits the last place to 3 choices

The first constraint limits the 1st place to 5 choices (repetition allowed)

*ps: see that in repetition being allowed, the 1st two constraints are independent

The third one says that the last one is dependent on the sum of the first three, so 2nd place we can choose from any 6 options. But, then the next one whatever be will have 2 choices. **explained below

So total cases would be 3*5*6*2=180 cases

**if till the third digit that we select, the sum is of the form 3k, then the fourth value should be of the form 3k (which are 2 possibilities here, viz. 0&3)

if till the third digit ..........of the form 3k+1, then the fourth value should be of the form 3k+2 (which are also 2 possibilities)

I think by now you should understand that there are 6 consecutive possibilities, and 6 is a multiple of 3

Logical/Observational approach

Firstly, you should know that this process would work only when repetition is allowed and few other constraints.

The lucky part is that the number of digits that we can use is actually 6. And more so, they are consecutive. So, here we can employ the third constraint of the previous solution and merge it up with 2nd.

So, for a four digit number ABCD, we are focussing on D

For a specific set of ABC, we would have 6 consecutive numbers

ABC0 to ABC5 and (number theory)/(logic) tells us that out of 6 consecutive numbers exactly 1 must be divisible by 6

So, our random chance of selection would be 1/6 of the total cases

Thus the answer should be (1/6)*5*6*6*6=180 cases