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# Six P’s have to be placed in the squares of a diagram given below such that each column contains at least one P. In how many ways can this be done?(a) 28(b) 26(c) 24(d) 30

Last updated date: 20th Jun 2024
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Hint: We solve this problem by using the permutations and combinations. First we fix one P in each column by selecting one column from each column which will be the permutations. Then, we place remaining P’s in the remaining squares which in turn the total number of ways by combining both conditions will be combinations. We have the formula that the number of ways of selecting $'r'$ boxes from $'n'$ boxes is given as ${}^{n}{{C}_{r}}$ where,
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$

We are given that there are 8 squares in the formation shown above.
We are given that there will be at least one P in each column.
Here, we can see that there are 3 columns.
Now, let us fix 3 P’s in each column.
Here, we can see that there are 2 boxes in the first column.
We know that the number of ways of selecting $'r'$ boxes from $'n'$ boxes is given as ${}^{n}{{C}_{r}}$ where,
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
By using the above formula we get the number of ways of selecting 1 box from 2 boxes in the first column as
\begin{align} & \Rightarrow {}^{2}{{C}_{1}}=\dfrac{2!}{1!\left( 2-1 \right)!} \\ & \Rightarrow {}^{2}{{C}_{1}}=2 \\ \end{align}
Now, we can see that there are 4 boxes in the second column.
Similarly, for the second column we get the number of ways of selecting 1 box from 4 boxes as
\begin{align} & \Rightarrow {}^{4}{{C}_{1}}=\dfrac{4!}{1!\left( 4-1 \right)!} \\ & \Rightarrow {}^{4}{{C}_{1}}=4 \\ \end{align}
Here, we can see that there are 2 boxes in the third column.
Similarly, for the third column we get the number of ways of selecting 1 box from 2 boxes as
\begin{align} & \Rightarrow {}^{2}{{C}_{1}}=\dfrac{2!}{1!\left( 2-1 \right)!} \\ & \Rightarrow {}^{2}{{C}_{1}}=2 \\ \end{align}
Let us assume that the number of ways of fixing one P in each column as $'x'$
Here, we know that the total number of ways of placing one P in each column will be the permutations for the number of ways of placing one P in the first, second and third columns.
By using the above condition we get the total number of ways of placing one P in each column as
\begin{align} & \Rightarrow x=2\times 4\times 2 \\ & \Rightarrow x=16 \\ \end{align}
So, we can say that we have placed 3 P’s in 3 boxes from each column.
Here, we can see that there are a total of 8 boxes.
So, we can say that there are 3 remaining of 3 P’s and 5 boxes.
Let us assume that the number of ways of placing the remaining P’s in remaining boxes as $'y'$
Now, we get the number of ways of placing 3 P’s in 5 boxes by selecting the 3 boxes from 5 boxes as
\begin{align} & \Rightarrow y={}^{5}{{C}_{3}} \\ & \Rightarrow y=\dfrac{5!}{3!\left( 5-3 \right)!} \\ & \Rightarrow y=10 \\ \end{align}
Now, let us assume that the total number of ways of placing 6 P’s in given 8 boxes as $'N'$
Here, we can see that the total number of ways of placing 6 P’s in given 8 boxes in the required condition is the combination of number of ways of placing one P in each column and number of ways of placing the remaining P’s in remaining boxes
Therefore, we get the required number of ways as
\begin{align} & \Rightarrow N=x+y \\ & \Rightarrow N=16+10 \\ & \Rightarrow N=26 \\ \end{align}
Therefore, the number of ways of placing 6 P’s in given 8 boxes such that each column has at least 1 P is 26.
So, option (b) is the correct answer.
So, the correct answer is “Option (b)”.

Note: Students may make mistakes in understanding the permutations and combinations.
The first part that is the number of ways of placing one P in each column will be the permutations of the number of ways of placing one P in first, second and third columns. So we get
\begin{align} & \Rightarrow x=2\times 4\times 2 \\ & \Rightarrow x=16 \\ \end{align}
Now, the final part that is the number of ways of placing 6 P’s in 8 boxes in the required condition is the combination of number of ways of placing one P in each column and number of ways of placing the remaining P’s in remaining boxes. So we get
$\Rightarrow N=x+y$
These parts need to be taken care of, which is permutations and which combinations are important in this problem.