Question:
Let $P$ and $Q$ are two sets such that
\[
n(P)=27,\quad n(Q)=17,\quad n(P\cap Q)=5.
\]
If $x$ is the number of ways of selecting $7$ elements from $P$ such that all the elements of $P\cap Q$ are in each selection and $y$ is the number of ways of selecting $10$ elements from $Q$ such that no element of $P\cap Q$ is present in any selection, then $x+y+1=$
Let $P$ and $Q$ are two sets such that
\[
n(P)=27,\quad n(Q)=17,\quad n(P\cap Q)=5.
\]
If $x$ is the number of ways of selecting $7$ elements from $P$ such that all the elements of $P\cap Q$ are in each selection and $y$ is the number of ways of selecting $10$ elements from $Q$ such that no element of $P\cap Q$ is present in any selection, then $x+y+1=$
Show Hint
For compulsory selections:
\[
\text{Required choices} - \text{Fixed elements}
\]
For prohibited selections:
\[
\text{Total elements} - \text{Restricted elements}
\]
Updated On: Jun 17, 2026
- $231$
- $248$
- $297$
- $298$
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Concept:
In combinations:
\[
^nC_r=\frac{n!}{r!(n-r)!}
\]
When some elements are compulsory, first include them and then choose the remaining elements.
When some elements are excluded, remove them from the total set before selection.
Step 1: Find $x$.
Given: \[ n(P)=27, \qquad n(P\cap Q)=5 \] All elements of \[ P\cap Q \] must be included in every selection. Thus, out of the required $7$ elements, $5$ are already fixed. Remaining elements to be selected: \[ 7-5=2 \] Available remaining elements in $P$: \[ 27-5=22 \] Hence: \[ x={}^{22}C_2 \] \[ =\frac{22\times21}{2} \] \[ =231 \]
Step 2: Find $y$.
Given: \[ n(Q)=17, \qquad n(P\cap Q)=5 \] No element of \[ P\cap Q \] is allowed. Thus usable elements from $Q$: \[ 17-5=12 \] We select $10$ elements from these $12$ elements. Hence: \[ y={}^{12}C_{10} \] \[ ={}^{12}C_2 \] \[ =\frac{12\times11}{2} \] \[ =66 \]
Step 3: Compute $x+y+1$.
\[ x+y+1 = 231+66+1 \] \[ =298 \] Hence, \[ \boxed{298} \]
Step 1: Find $x$.
Given: \[ n(P)=27, \qquad n(P\cap Q)=5 \] All elements of \[ P\cap Q \] must be included in every selection. Thus, out of the required $7$ elements, $5$ are already fixed. Remaining elements to be selected: \[ 7-5=2 \] Available remaining elements in $P$: \[ 27-5=22 \] Hence: \[ x={}^{22}C_2 \] \[ =\frac{22\times21}{2} \] \[ =231 \]
Step 2: Find $y$.
Given: \[ n(Q)=17, \qquad n(P\cap Q)=5 \] No element of \[ P\cap Q \] is allowed. Thus usable elements from $Q$: \[ 17-5=12 \] We select $10$ elements from these $12$ elements. Hence: \[ y={}^{12}C_{10} \] \[ ={}^{12}C_2 \] \[ =\frac{12\times11}{2} \] \[ =66 \]
Step 3: Compute $x+y+1$.
\[ x+y+1 = 231+66+1 \] \[ =298 \] Hence, \[ \boxed{298} \]
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