An urn contains 5 red marbles,
Question:

An urn contains 5 red marbles, 4 black marbles and 3 white marbles. Then the number of ways in which 4 marbles can be drawn so that at the most three of them are red is_________.

Solution:

0 Red, 1 Red, 2 Red, 3 Red

Number of ways of selecting atmost three red balls

$={ }^{7} C_{4}+{ }^{5} C_{1} \cdot{ }^{7} C_{3}+{ }^{5} C_{2} \cdot{ }^{7} C_{2}+{ }^{5} C_{3} \cdot{ }^{7} C_{1}$

$=35+175+210+70=490$

Administrator

Leave a comment

Please enter comment.
Please enter your name.