JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Permutations and Combinations

With at least 4 from each of sections A(8), B(6), C(6), the base allocation $4+4+4=12$ leaves 3 more marks to distribute, subject to A allowing up to 4 extra (cap 8) and B, C allowing up to 2 extra each (cap 6).

Writing the extra amounts as $a^{\prime },b^{\prime },c^{\prime }\ge 0$ with $a^{\prime }+b^{\prime }+c^{\prime }=3$, $b^{\prime }\le 2$, $c^{\prime }\le 2$: among the 10 unrestricted non-negative solutions, only $(0,3,0)$ and $(0,0,3)$ violate the caps, leaving 8 valid distributions, giving $(a,b,c)$ as
$(7,4,4),(6,5,4),(6,4,5),(5,6,4),(5,4,6),(4,6,5),(4,5,6),(5,5,5)$.

Computing $\left(\genfrac{}{}{0ex}{}{8}{a}\right)\left(\genfrac{}{}{0ex}{}{6}{b}\right)\left(\genfrac{}{}{0ex}{}{6}{c}\right)$ for each:

$(7,4,4)$: $8\cdot 15\cdot 15=1800$
$(6,5,4)$: $28\cdot 6\cdot 15=2520$
$(6,4,5)$: $28\cdot 15\cdot 6=2520$
$(5,6,4)$: $56\cdot 1\cdot 15=840$
$(5,4,6)$: $56\cdot 15\cdot 1=840$
$(4,6,5)$: $70\cdot 1\cdot 6=420$
$(4,5,6)$: $70\cdot 6\cdot 1=420$
$(5,5,5)$: $56\cdot 6\cdot 6=2016$

Adding: $1800+2520+2520+840+840+420+420+2016=11376$.