A-BILLI : Creating Problems on me ?

Since there are 6 letters, one row must have 2 letters on it and there are 3 possible rows for the placement of those two letters.

First let's choose 6 boxes for the letters. The number of ways to do this is (number of rows with 3 boxes) x (number of ways to choose two boxes from a row of three boxes) x (number of ways to choose a box from a row of three boxes) = 3 x 3 x 3^2 (squared since we must get one box each from the other 2 rows) = 81

Next step is finding the number of ways to arrange the six letters into these 6 boxes that we chose earlier. The number of ways to do this is 6!/(2! x 2!) (because there are 2 I's and 2 L's) = 180

Therefore to get the number of distinct ways to place the 6 letters in the figure such that no two row remains empty is just the product of the two numbers we get which is 81 x 180 = 14580

There are two ways of doing combination problems

1. By adding different possibilities taking into account the constraints

2. By taking the total and subtracting what is not possible

I have done it by first method

There are 5 rows and 6 letters so one row will have 2 letters and others 1 letter each

A, B - Dissimilar Letters

I, I and L,L - Similar Letters

Case I: Two letters which come in the same row are similar (either I, I or L,L)

L1= 2 (Selecting first two) $\times$ 3 (Arranging first two) $\times$ 3 $\times$ 3 (Selecting places for remaining four) $\times$ $\frac{4!}{2!}$ (Arranging the remaining 4)

Case II: Two Letters which come in the same row are dissimilar (A, B)

L2= $1\times3\times2\times3\times3\times \frac{4!}{2!2!}$

Case III: Two Letters which come in the same row are 1 similar and other dissimilar (one from A,B and other from I,L)

L3 = $2\times2\times3\times2\times3\times3\times \frac{4!}{2!}$

Case IV: Two Letters which come in the same row are dissimilar but from the similar group (I and L)

L4= $1\times3\times2\times3\times3\times4!$

Add all and multiply by 3 since there are three rows where 2 letters are possible

Total = $3 \times (L1+L2+L3+L4)$

         = 14580