Question

Assertion: If \[p,q < r\], then the number of selections of \[p+q\] things taking \[r\] at a time, where \[p\] things are identical and other \[q\] things are identical is, \[p+q-r+1\]
Reason: If \[p,q > r\], then the number of selections of \[p+q\] things taking \[r\] at a time, where \[p\] things are identical and other \[q\] things are identical is, \[r-1\]
(a) Both assertion and reason are correct and reason is the correct explanation of assertion.
(b) Both assertion and reason are correct but reason is not correct explanation of assertion
(c) Assertion is correct but reason is wrong
(d) Assertion is wrong but reason is correct

Answer 1

Hint: We solve this problem by taking the number of possibilities of selecting the things from the set of \[p\] identical things and the set of \[q\] identical things.
We use the condition that selection of \[r\] things from \[n\] identical things can be done only in 1 way.
By this condition we can say that the total number of selections will be the number of ways of selecting \[r\] things from a set of \[p\] identical things and a set of \[q\] identical things.
We calculate the number of ways for each assertion and reason.

Complete step by step answer:
Let us solve for assertion first that is \[p,q < r\]
We are given that we need to select \[p+q\] things taking \[r\] at a time, where \[p\] things are identical and other \[q\] things are identical
Let us assume that there are a total of \[k\] possibilities such that we select \[r\] things from a set of \[p\] identical things and a set of \[q\] identical things.
We know that the condition that selection of \[r\] things from \[n\] identical things can be done only in 1 way.
By the above condition we can say that the total number of ways of selecting \[p+q\] things taking \[r\] at a time, where \[p\] things are identical and other \[q\] things are identical is \[k\]
Now, let us create a table such that it contains the possibility number, things selected from \[p\] identical things and things selected from \[q\] identical things as follows

Possibility number	Things selected from \[p\] identical things	Things selected from \[q\] identical things
1	\[p\]	\[r-p\]
2	\[p-1\]	\[r-p+1\]
3	\[p-2\]	\[r-p+2\]
..	..	..
…	…	…
k	\[x\]	\[q\]

Here, we can see that the last possibility will come when there are \[q\] things selected from \[q\] identical things
We know that in each possibility we need to get \[r\] things at a time
By using the above condition to last possibility we get
\[\Rightarrow x+q=r.......equation(i)\]
Now, let us rewrite the expressions of second column in the above table in terms of its possibility number then we get

Possibility number	Things selected from \[p\] identical things	Things selected from \[q\] identical things
1	\[\left( p+1 \right)-1\]	\[r-p\]
2	\[\left( p+1 \right)-2\]	\[r-p+1\]
3	\[\left( p+1 \right)-3\]	\[r-p+2\]
..	..	..
…	…	…
k	\[\left( p+1 \right)-k\]	\[q\]

Here we can see that the value of \[x\] is given as
\[\Rightarrow x=p+1-k\]
By substituting this value in equation (i) we get
\[\begin{align}
  & \Rightarrow p+1-k+q=r \\
 & \Rightarrow k=p+q-r+1 \\
\end{align}\]
Therefore, we can conclude that the total number of selections in this case is \[p+q-r+1\]
Now, let us solve for reason statement that is \[p,q>r\]
We are given that we need to select \[p+q\] things taking \[r\] at a time, where \[p\] things are identical and other \[q\] things are identical
Let us assume that there are a total of \[k\] possibilities such that we select \[r\] things from a set of \[p\] identical things and a set of \[q\] identical things.
By the condition of selecting things from identical thing, we can say that the total number of ways of selecting \[p+q\] things taking \[r\] at a time, where \[p\] things are identical and other \[q\] things are identical is \[k\]
Now, let us create a table such that it contains the possibility number, things selected from \[p\] identical things and things selected from \[q\] identical things as follows

Possibility number	Things selected from \[p\] identical things	Things selected from \[q\] identical things
1	\[r\]	\[0\]
2	\[r-1\]	\[1\]
3	\[r-2\]	\[2\]
..	..	..
…	…	…
k	\[0\]	\[r\]

Here, we can see that the last possibility will come when there are \[r\] things selected from \[q\] identical things
We know that in each possibility we need to get \[r\] things at a time
Now, let us rewrite the expressions of second column in the above table in terms of its possibility number then we get

Possibility number	Things selected from \[p\] identical things	Things selected from \[q\] identical things
1	\[\left( r+1 \right)-1\]	\[0\]
2	\[\left( r+1 \right)-2\]	\[1\]
3	\[\left( r+1 \right)-3\]	\[2\]
..	..	..
…	…	…
k	\[\left( r+1 \right)-\left( r+1 \right)\]	\[r\]

Here, by comparing the first and second columns we can say that
\[\Rightarrow k=r+1\]
Therefore, we can conclude that the total number of selections in this case is \[r+1\]

So, we can say that the assertion is correct but reason is wrong.
So, the correct answer is “Option c”.

Note: Students may make mistakes in the selection of identical things.
We have the formula that number of ways of selecting \[r\] things from \[n\] things is given as \[{}^{n}{{C}_{r}}\]
Here, this formula is applicable when the \[n\] things are different. But the case changes when the \[n\] things are identical.
We use the condition that selection of \[r\] things from \[n\] identical things can be done only in 1 way.
This is because the selection of any \[r\] things from the \[n\] identical things is the same because we cannot differentiate the things in each case. As all the cases have the same items selected, that is considered only 1 way.