given piles of identical red, blue and green balls where each pile contains at least 10 balls, how many ways can 10 balls be selected such that exactly two green balls are selected and at most one red ball is selected?
indicate your subject instead of just a d!!
To solve this problem, we can use combinatorics - specifically, the concept of combinations.
First, let's break down the problem:
- We need to select exactly two green balls.
- We need to select at most one red ball.
- We need to select 10 balls in total.
To find the number of ways to select the balls, we'll consider different cases:
Case 1: Selecting two green balls and no red ball
In this case, we have to select 8 balls from the remaining red and blue piles. Since each pile contains at least 10 balls, we can consider the red and blue piles together as one pile. So, we have one pile of (10-2=) 8 balls. Using combinations, we can choose 8 balls from this pile in (8 choose 8) ways, which is 1.
Case 2: Selecting two green balls and one red ball
In this case, we need to select 7 balls from the remaining piles. Again, we can consider the red and blue piles together. So, we have one pile of (10-2-1=) 7 balls. We can choose 7 balls from this pile in (7 choose 7) ways, which is 1.
Case 3: Selecting two green balls and two red balls
In this case, we need to select 6 balls from the remaining piles. Now, we have two piles: the red pile and the blue pile. We can choose 6 balls from the two piles in (6 choose 2) ways, which is 15.
Therefore, the total number of ways to select the balls that satisfy the given conditions is 1 + 1 + 15 = 17.