1. **State the problem:** Calculate the value of $\frac{{24C2 \times 24C2}}{{52C4}}$ where $nCr$ denotes the combination formula $\binom{n}{r}$. 2. **Recall the combination formula:** $$\binom{n}{r} = \frac{n!}{r!(n-r)!}$$ This formula gives the number of ways to choose $r$ items from $n$ without regard to order. 3. **Calculate each combination:** $$24C2 = \binom{24}{2} = \frac{24!}{2! \times 22!} = \frac{24 \times 23}{2} = 276$$ 4. **Calculate $52C4$:** $$52C4 = \binom{52}{4} = \frac{52!}{4! \times 48!} = \frac{52 \times 51 \times 50 \times 49}{4 \times 3 \times 2 \times 1}$$ Calculate numerator: $$52 \times 51 = 2652$$ $$2652 \times 50 = 132600$$ $$132600 \times 49 = 6497400$$ Calculate denominator: $$4 \times 3 \times 2 \times 1 = 24$$ So, $$52C4 = \frac{6497400}{24} = 270725$$ 5. **Substitute values into the original expression:** $$\frac{24C2 \times 24C2}{52C4} = \frac{276 \times 276}{270725} = \frac{76176}{270725}$$ 6. **Simplify the fraction if possible:** Check for common factors; 76176 and 270725 share no obvious common factors, so the fraction is simplified. **Final answer:** $$\frac{24C2 \times 24C2}{52C4} = \frac{76176}{270725} \approx 0.2813$$