A box contains 8 red balls, 7 blue balls, and 6 orange balls. In how many ways can 6 balls be chosen if there should be 2 balls each color?
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A box contains 8 red balls, 7 blue balls, and 6 orange balls. In how many ways can 6 balls be chosen if there should be 2 balls each color?
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Verified answer
In the given problem, apply your knowledge in algebraic expression and basic mathematics especially in solving word problem about combinatorics.
Problem
A box contains 8 red balls, 7 blue balls, and 6 orange balls. In how many ways can 6 balls be chosen if there should be 2 balls each color?
Given
8 red balls
7 blue balls
6 orange balls
Solution
Use the formula for combination, nCr = n!/r!(n-r)!
If the restriction is 2 balls each color
→ nCr = nCr (red) × nCr (blue) × nCr (orange)
nCr = 8C2 × 7C2 × 6C2
nCr = 8!/2!(8-2)! × 7!/2!(7-2)! × 6!/2!(6-2)!
nCr = 28 × 21 × 15
nCr = 8 820
Answer
8 820 ways
Learn more about combination here at brainly.ph/question/13828020
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