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 $x+y+1=$

• A. $231$
• B. $248$
• C. $297$
• D. $298$

Hint:

For compulsory selections: \[ \text{Required choices} - \text{Fixed elements} \] For prohibited selections: \[ \text{Total elements} - \text{Restricted elements} \]

Answer 1

The Correct Option is D

Step 1: Read the set sizes.
We have $n(P)=27$, $n(Q)=17$, and $n(P\cap Q)=5$. We must find two counts and then add them with $1$.
Step 2: Build the count $x$.
For $x$ we pick $7$ elements from $P$ that must include all $5$ common elements. So $5$ are forced and we need $2$ more.
Step 3: Choose the remaining for $x$.
The non-common part of $P$ has $27-5=22$ elements. Choose $2$ of them: $x={}^{22}C_2=\dfrac{22\times21}{2}=231$.
Step 4: Build the count $y$.
For $y$ we pick $10$ elements from $Q$ with none of the $5$ common ones. So we may only use $17-5=12$ elements.
Step 5: Choose for $y$.
Choose $10$ from $12$: $y={}^{12}C_{10}={}^{12}C_2=\dfrac{12\times11}{2}=66$.
Step 6: Add as asked.
Then $x+y+1=231+66+1=298$. \[ \boxed{298} \]