Question
Among a set of 4 black balls and 3 white balls, how
many selections of 3 balls can be made such that at least 2 of them are black balls?Solution
Selecting at least 2 black balls from a set of 4 black balls in total selection of 3 balls can be 2 black and 1 white balls 3 black and 0 white balls Therefore, our solution expression looks 4C2 × 3C1 + 4C3 × 3C0 6 X 3 + 4 X 1 = 18 + 4 = 22 ways
More Permutation and combination Questions
What will come in the place of question mark (?) in the given expression?
{(3600 ÷ 20 of 18) × 30 + 240} = ? % of 1200
1.25 × 36 + 2.75 × 40 = ? × 3.1
72% of 486 – 64% of 261 = ?
(49 x ?) / 3.5 + 389 = 627
(25.111 % of 200) × 26 ÷ 12.99 – 18.88 × 15.82 + 150.33% of 3√ 4917 = ? – 200
...72.5% of 400 – 23.25% of 1020 = 105% of ?
(3984 ÷ 24) x (5862 ÷ 40) = ?
154 × 7 + 480 × 5 =?% of 6956
40% of (362 ÷ 0.05) = ?
(392 + 427 + 226 – 325) ÷ (441 + 128 – 425) = ?Â
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