Math JEE Adv. 2025 paper 1

 

To determine the number of reflexive and symmetric relations $R$ on the set $\{a,b,c,d,e,f\}$ that contain exactly 10 elements, we follow these steps:

Step 1: Understand Reflexive and Symmetric Relations

• Reflexive: The relation must include all pairs $(x,x)$ for each $x\in \{a,b,c,d,e,f\}$. There are 6 such pairs.

• Symmetric: If $(x,y)\in R$, then $(y,x)\in R$ must also be included. Each unordered pair $\{x,y\}$ (where $x\mathrm{\ne }y$) corresponds to two ordered pairs $(x,y)$ and $(y,x)$.

Step 2: Count the Total Possible Pairs

• The total number of possible ordered pairs in $\{a,b,c,d,e,f\}\times \{a,b,c,d,e,f\}$ is $6\times 6=36$.

• Out of these, 6 are reflexive pairs $(x,x)$.

• The remaining $36-6=30$ pairs are non-reflexive, which come in symmetric pairs $(x,y)$ and $(y,x)$. Thus, there are $15$ unique unordered pairs.

Step 3: Determine the Number of Relations

• A reflexive relation must include all 6 reflexive pairs.

• To have a total of 10 elements, the relation must include $10-6=4$ additional non-reflexive pairs. Since the relation is symmetric, these must be added as symmetric pairs $(x,y)$ and $(y,x)$.

• Each such addition counts as 2 elements, so we need to choose $2$ unordered pairs (since $2\times 2=4$ elements).

Step 4: Calculate the Number of Ways to Choose Pairs

• The number of ways to choose 2 unordered pairs from the 15 available is:

  $$<span\; style="background-color:\; white;">(</span>\genfrac{}{}{0px}{}{\mathrm{<span\; style="background-color:\; white;">15</span>}}{\mathrm{<span\; style="background-color:\; white;">2</span>}}<span\; style="background-color:\; white;">)</span><span\; style="background-color:\; white;">=</span>\mathrm{<span\; style="background-color:\; white;">105.</span>}$$

Final Answer

The number of such relations $R$ is $\boxed{{\displaystyle 105}}$.