Q48 / 75 · JEE Main 2026 (22 January Shift 1) · Mathematics
Let ABC be a triangle. Consider four points $\mathrm{p}_{1}, \mathrm{p}_{2}, \mathrm{p}_{3}, \mathrm{p}_{4}$ on the side AB, five points $p_{5}, p_{6}, p_{7}, p_{8}, p_{9}$ on the side $B C$, and four points $p_{10}, p_{11}, p_{12}, p_{13}$ on the side AC. None of these points is a vertex of the triangle ABC. Then the total number of pentagons, that can be formed by taking all the vertices from the points $\mathrm{p}_{1}, \mathrm{p}_{2}, \ldots, \mathrm{p}_{13}$, is $\_\_\_\_$
Correct answer: 660
Solution
Points: 4 on AB, 5 on BC, 4 on AC. For a valid pentagon, no 3 vertices can be collinear, so at most 2 points from each side.
Only valid distribution $(a, b, c)$ with $a+b+c = 5$ and each $\le 2$ is permutations of $(1, 2, 2)$:
$(1,2,2)$: $\binom{4}{1}\binom{5}{2}\binom{4}{2} = 4 \cdot 10 \cdot 6 = 240$
$(2,1,2)$: $\binom{4}{2}\binom{5}{1}\binom{4}{2} = 6 \cdot 5 \cdot 6 = 180$
$(2,2,1)$: $\binom{4}{2}\binom{5}{2}\binom{4}{1} = 6 \cdot 10 \cdot 4 = 240$
Total $= 240 + 180 + 240 = 660$.
Only valid distribution $(a, b, c)$ with $a+b+c = 5$ and each $\le 2$ is permutations of $(1, 2, 2)$:
$(1,2,2)$: $\binom{4}{1}\binom{5}{2}\binom{4}{2} = 4 \cdot 10 \cdot 6 = 240$
$(2,1,2)$: $\binom{4}{2}\binom{5}{1}\binom{4}{2} = 6 \cdot 5 \cdot 6 = 180$
$(2,2,1)$: $\binom{4}{2}\binom{5}{2}\binom{4}{1} = 6 \cdot 10 \cdot 4 = 240$
Total $= 240 + 180 + 240 = 660$.
Hence, the correct answer is 660.