JEE Main 2026 — 4 April, Morning Shift — Question 43

$\mathrm{x}_{1} \mathrm{x}_{2} \mathrm{x}_{3} \mathrm{x}_{4} \mathrm{x}_{5} \mathrm{x}_{6} \mathrm{x}_{7} \mathrm{x}_{8}$ Exactly 4 heads in first six Tosses. $\left(\mathrm{x}_{1} \mathrm{x}_{2} \mathrm{x}_{3} \mathrm{x}_{4} \mathrm{x}_{5} \mathrm{x}_{6}\right)$ Exactly 3 heads in last 5 Tosses. i.e. $\left(\mathrm{x}_{4} \mathrm{x}_{5} \mathrm{x}_{6} \mathrm{x}_{7} \mathrm{x}_{8}\right)$ Common tosses $\left(\mathrm{x}_{4} \mathrm{x}_{5} \mathrm{x}_{6}\right)$ Let k be number of heads on $\left(\mathrm{x}_{4} \mathrm{x}_{5} \mathrm{x}_{6}\right)$ $\mathrm{k}=1 \therefore$ ways to overlap $={ }^{3} \mathrm{C}_{1}$ ${ }^{3} \mathrm{C}_{3} \times{ }^{3} \mathrm{C}_{1} \times{ }^{2} \mathrm{C}_{2}=3$ ways $\mathrm{k}=2$ ${ }^{3} \mathrm{C}_{2} \times{ }^{3} \mathrm{C}_{2} \times{ }^{2} \mathrm{C}_{1}=18$ $\mathrm{k}=3$ ${ }^{3} \mathrm{C}_{1} \times 1=3$ Total ways $24\left(\frac{1}{2}\right)^{8}$ $96 \times \frac{24}{28}=\frac{6 \times 24}{16}=9$