Question

Find the number of ways in which AAABBB can be placed in the squares of figure as shown, so that no row remains empty?

Answer 1

First find the number of ways to select the boxes in which the given letters are to be filled. Select any one row in which three boxes are there by using the relation ${}^{3}{{C}_{1}}$. Select any two boxes of this row to be filled by 2 letters by using the formula ${}^{3}{{C}_{2}}$. Now, select any one box of the remaining two rows which is to be filled with 1 letter each by using the formula ${}^{3}{{C}_{1}}$ for each. Multiply all these relations to get the total number of ways of selection of boxes. Finally, arrange six letters in the six boxes by using the relation $\dfrac{6!}{3!\times 3!}$ as both the letters are repeating three times each. Take the product of the number of arrangements with the total ways of selection of the boxes to get the answer.

Complete step by step answer:
Here we have been provided with the letters AAABBB and we are asked to arrange them in the given structure such that no row remains empty.

Now, we can see that there are 5 rows in which two rows have 1 box each and 3 rows have 3 boxes each. Since, no row should be left empty so the two rows that contain one box each must be filled and there is only one way to select them that means we need not to think about them.
The three rows that contain 3 boxes each can be filled in several ways. Any one of these three rows must be filled with 2 letters and the remaining two rows with a letter. So, let us find the number of ways in which the boxes can be selected.
Number of ways to select 1 row among the 3 rows that should be filled with 2 letters = ${}^{3}{{C}_{1}}$.
Number of ways to select 2 boxes to be filled with 2 letters among the 3 boxes = ${}^{3}{{C}_{2}}$.
Number of ways to select 1 box to be filled with 1 letter each of the 2 remaining rows containing 3 boxes each = ${}^{3}{{C}_{1}}\times {}^{3}{{C}_{1}}$.
Therefore, total number of ways to select 6 boxes to be filled with 6 letters = ${}^{3}{{C}_{1}}\times {}^{3}{{C}_{2}}\times {}^{3}{{C}_{1}}\times {}^{3}{{C}_{1}}=81$
Now, we need to arrange 6 letters in 6 boxes that are selected. If all the letters were different then the number of ways for the arrange would have been $6!$ but here we can see that both the letters A and B are repeating 3 times each so the effective arrangements will be $\dfrac{6!}{3!\times 3!}$.
Therefore, the total number of ways in which the letters can be filled so that no row remains empty will be = $81\times \dfrac{6!}{3!\times 3!}=1620$.
Hence, the total number of ways will be 1620.

Note: Remember the combination formula ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ which is used when we select r things out of n things. Note that here we do not have to consider the sum of the ways of selection and arrangement but we have to consider their product. This is because we have to perform both the functions. If we would have to perform only one function then in such cases we consider the sum instead of product.