A box contains 5 red balls and 6 black balls. in how many ways can 6 balls be selected so that there are at least two balls of each color?
i) 2 red,4 black => C(5,2)*C(6,4) =150
ii)3 red,3 black => C(5,3)*C(6,3) =200
iii)4red,2 black => C(5,4)*C(6,2) =75
Therefore total number of ways =
150+200+75 = 425 ways
Hope this helps! Cheers! :)
To find the number of ways to select 6 balls from a box containing 5 red balls and 6 black balls, such that there are at least two balls of each color, we need to consider two cases:
Case 1: Select exactly 2 red balls and 4 black balls.
In this case, we have to choose 2 red balls from the 5 available and 4 black balls from the 6 available. This can be done using the combination formula.
Number of ways to select 2 red balls from 5 = C(5, 2) = 10
Number of ways to select 4 black balls from 6 = C(6, 4) = 15
Therefore, the number of ways to select exactly 2 red balls and 4 black balls is 10 * 15 = 150.
Case 2: Select exactly 3 red balls and 3 black balls.
In this case, we have to choose 3 red balls from the 5 available and 3 black balls from the 6 available.
Number of ways to select 3 red balls from 5 = C(5, 3) = 10
Number of ways to select 3 black balls from 6 = C(6, 3) = 20
Therefore, the number of ways to select exactly 3 red balls and 3 black balls is 10 * 20 = 200.
Finally, we add the results from both cases to get the total number of ways to select 6 balls with at least two balls of each color:
Total number of ways = 150 + 200 = 350
So, there are 350 ways to select 6 balls such that there are at least two balls of each color.