Q57 / 160 · AP EAMCET 2021 (19 Aug Shift 2) · Mathematics

In how many ways $4$ balls can be picked from $6$ black and $4$ green colored balls such that at least one black ball is selected?

Solution

All possible cases if at least one Black ball is selected are,

$1 B 3 G, 2 B 2 G, 3 B 1 G, 4 B 0 G$

Then total number of ways are,

$= {}^{6}{C_{1}} \times {}^{4}{C_{3}} + {}^{6}{C_{2} \times} {}^{4}C_{2} + {}^{6}{C_{3} \times} {}^{4}{C_{1}} + {}^{6}{C_{4}}$

$= 6 \times 4 + 15 \times 6 + 20 \times 4 + 15$

$= 209$

Hence, the correct answer is option: 'C'.