JEE Advanced 2024 — Paper 1 — Question 11

Case-1: $\mathrm{S}_{2}$ is in team X and $\mathrm{S}_{1}$ is in team $\mathrm{Y},{ }^{7} \mathrm{C}_{1} \times{ }^{6} \mathrm{C}_{2} \times{ }^{4} \mathrm{C}_{4}=105$

Case-2: $\mathrm{S}_{2}$ is in team X but $\mathrm{S}_{1}$ is not in team $\mathrm{Y},{ }^{7} \mathrm{C}_{2} \times{ }^{5} \mathrm{C}_{2} \times{ }^{3} \mathrm{C}_{3}=140$

Case-3: $\mathrm{S}_{2}$ is not in team X but $\mathrm{S}_{1}$ is in team $\mathrm{Y},{ }^{7} \mathrm{C}_{2} \times{ }^{5} \mathrm{C}_{2} \times{ }^{3} \mathrm{C}_{3}=210$

Case-4: $\mathrm{S}_{2}$ is not in team X and $\mathrm{S}_{1}$ is not in team $\mathrm{Y},{ }^{7} \mathrm{C}_{2} \times{ }^{5} \mathrm{C}_{3} \times{ }^{2} \mathrm{C}_{2}=210$

$\therefore$ Total ways to form such teams $=105+140+210+210=665$