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Question: The number of ways in which the letters of the word 'PERSON' can be placed in the square of the give...

The number of ways in which the letters of the word 'PERSON' can be placed in the square of the given figure so that no row remains empty is

A. 24×6!24 \times 6!
B. 26×6!26 \times 6!
C. 26×7!26 \times 7!
D. 27×6!27 \times 6!

Explanation

Solution

We can see that the word 'PERSON' has 6 different letters. So, we will have to select 6 squares, taking at least one from each row and then arranging in each selection. Also, we will use the concept that selection of r things from n different things is equal to nCr{}^n{C_r}.

Complete step-by-step solution:
We have been given a word 'PERSON' and asked to place the letters of the word in the square of the given figure so that no row remains empty.

Let us first select places in each row given that no row remains empty.
Case I:
When selection of places from R3=4 , R2=1 and   R1=1{R_3} = 4\,,\,{R_2} = 1\,and\,\,\,{R_1} = 1
Number of selection =4C4×2C1×2C1 = {}^4{C_4} \times {}^2{C_1} \times {}^2{C_1}
=1×2×2 =4\begin{array}{l} = 1 \times 2 \times 2\\\ = 4\end{array}
Case II:
When selection of places from R3=3 , R2=2 and   R1=1{R_3} = 3\,,\,{R_2} = 2\,and\,\,\,{R_1} = 1
Number of selection =4C3×2C2×2C1 = {}^4{C_3} \times {}^2{C_2} \times {}^2{C_1}
=4×1×2 =8\begin{array}{l} = 4 \times 1 \times 2\\\ = 8\end{array}
Case III:
When selection of places from R3=3 , R2=1 and   R1=2{R_3} = 3\,,\,{R_2} = 1\,and\,\,\,{R_1} = 2
Number of selection =4C3×2C1×2C2 = {}^4{C_3} \times {}^2{C_1} \times {}^2{C_2}
=4×2×1 =8\begin{array}{l} = 4 \times 2 \times 1\\\ = 8\end{array}
Case IV:
When selection of places from R3=2 , R2=2 and   R1=2{R_3} = 2\,,\,{R_2} = 2\,and\,\,\,{R_1} = 2
Number of selection =4C2×2C2×2C2 = {}^4{C_2} \times {}^2{C_2} \times {}^2{C_2}
=6×1×1 =6\begin{array}{l} = 6 \times 1 \times 1\\\ = 6\end{array}
So, total number of selection of 6 squares is 4+8+8+6=264 + 8 + 8 + 6 = 26
Now, for each selection of 6 squares, the number of arrangements of 6 letters is 6!6!.
Required number of ways =26×6! = 26 \times 6!
Therefore, the correct option is B.

Note: We cannot take the case when R3=3{R_3} = 3 and R2=3{R_2} = 3 . Since, we have to choose the 6 square place but it’s not possible in these cases. Sometimes we just forget to take the case when R3=3 , R2=1 and   R1=2{R_3} = 3\,,\,{R_2} = 1\,\text{and}\,\,\,{R_1} = 2 of selection of places because it is similar to the case when R3=3 , R2=2 and   R1=1{R_3} = 3\,,\,{R_2} = 2\,\text{and}\,\,\,{R_1} = 1. So, be careful while solving.