How many Number of 7digit natural number in which both 2&3 arepresent as digits if no 2`sare consecutive in any number?

Dear student

I’ll presume you mean 7-digit natural numbers with only 2s and 3s, but with no 2s adjacent to each other.

For example, 2323232 would be fine, but 3223332 would not be.

One way to do this is with a recurrence relation.

Let A(n) = the number of n-digit natural numbers with only 2s and 3s, but with no 2s adjacent to each other.

Consider the last digit. Either it’s a 2 or a 3. If it’s a 3, then the first n-1 digits can be any valid number, because any valid number followed by a 3 is valid. By definition, there are A(n-1) of these.

But the last digit could also be a 2. In this case, the n-1 digit could not be a 2, otherwise you’d have two 2s together. So the n-1 digit must be a 3. Now, the first n-2 digits can be any valid number. By definition, there are A(n-2) of these.

So this gives us the recurrence relation A(n) = A(n-1) + A(n-2). We’ll need two boundary conditions, since this is a 2nd order recurrence relation.

A(1) = 2, because either 2 or 3 is valid. A(2) = 3 because 23, 32, and 33 are valid 2-digit numbers.

Now we can use these to compute A(7), which is what we want. So:

A(3) = A(2) + A(1) = 3 + 2 = 5

A(4) = A(3) + A(2) = 5 + 3 = 8

A(5) = A(4) + A(3) = 8 + 5 = 13

A(6) = A(5) + A(4) = 13 + 8 = 21 and finally

A(7) = A(6) + A(5) = 21 + 13 = 34

And that’s our answer.

Regards

Arun (askIITians forum expert)