Let $$𝑁$$ be the number of nine-digit integers that can be obtained by permuting the digits of $$223334444$$ and which have at least one $$3$$ to the right of the right-most occurrence of $$4$$. What is the remainder when $$𝑁$$ is divided by $$100$$?

We have to arrange the multiset $$\{2,2,3,3,3,4,4,4,4\}$$ (two 2’s, three 3’s and four 4’s) in a row of nine positions so that at least one $$3$$ appears to the right of the right-most $$4$$.

Step 1 : Fix the positions of the two 2’s.
Choose any 2 of the 9 places for the identical 2’s.
Number of ways = $$\binom{9}{2}=36$$.

Step 2 : Work inside the remaining 7 positions.
After removing the two 2’s, seven places are left and must be filled with three 3’s and four 4’s.

The required condition “there is a 3 to the right of the right-most 4” means that, within these seven places, the last (right-most) of them must be a 3. Fix that last place as 3. The other six places have to accommodate the remaining
  • two 3’s
  • four 4’s

Number of distinct arrangements of those six symbols = $$\dfrac{6!}{2!\,4!}=15$$.

Step 3 : Combine the choices.
For each of the 36 ways to place the 2’s, there are 15 admissible ways to arrange the 3’s and 4’s.
Hence $$N = 36 \times 15 = 540$$.

Step 4 : Find the remainder mod 100.
$$540 \bmod 100 = 40$$.

Therefore the required remainder is 40.