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To solve this problem, we can use a combinatorial approach. We'll break it down into cases:
Choose 1 red ball and 1 blue ball from each of the 4 boxes.
Choose 2 red balls and 1 blue ball from one box, and 1 red ball and 2 blue balls from each of the remaining 3 boxes.
Choose 1 red ball and 2 blue balls from one box, and 2 red balls and 1 blue ball from each of the remaining 3 boxes.
Let's calculate the number of ways for each case:
Choosing 1 red ball and 1 blue ball from each of the 4 boxes:
(
3
1
)
×
(
2
1
)
(
1
3
)×(
1
2
) ways from each box
Total ways =
(
3
1
)
×
(
2
1
)
(
1
3
)×(
1
2
)^4
Choosing 2 red balls and 1 blue ball from one box, and 1 red ball and 2 blue balls from each of the remaining 3 boxes:
(
3
2
)
×
(
2
1
)
(
2
3
)×(
1
2
) ways to choose from one box
(
3
1
)
×
(
2
2
)
(
1
3
)×(
2
2
) ways to choose from each of the remaining 3 boxes
Total ways =
(
3
2
)
×
(
2
1
)
×
(
3
1
)
×
(
2
2
)
(
2
3
)×(
1
2
)×(
1
3
)×(
2
2
)^4
Choosing 1 red ball and 2 blue balls from one box, and 2 red balls and 1 blue ball from each of the remaining 3 boxes:
(
3
1
)
×
(
2
2
)
(
1
3
)×(
2
2
) ways to choose from one box
(
3
2
)
×
(
2
1
)
(
2
3
)×(
1
2
) ways to choose from each of the remaining 3 boxes
Total ways =
(
3
1
)
×
(
2
2
)
×
(
3
2
)
×
(
2
1
)
(
1
3
)×(
2
2
)×(
2
3
)×(
1
2
)^4
Now, we sum up the total number of ways from each case to get the final answer. Let's calculate it.
Let's calculate the number of ways for each case:
Choosing 1 red ball and 1 blue ball from each of the 4 boxes:
(
3
1
)
(
1
3
) ways to choose 1 red ball from each box
(
2
1
)
(
1
2
) ways to choose 1 blue ball from each box
Total ways =
(
3
1
)
×
(
2
1
)
(
1
3
)×(
1
2
)^4
Choosing 2 red balls and 1 blue ball from one box, and 1 red ball and 2 blue balls from each of the remaining 3 boxes:
(
3
2
)
(
2
3
) ways to choose 2 red balls from one box
(
2
1
)
(
1
2
) ways to choose 1 blue ball from one box
(
3
1
)
(
1
3
) ways to choose 1 red ball from each of the remaining 3 boxes
(
2
2
)
(
2
2
) ways to choose 2 blue balls from each of the remaining 3 boxes
Total ways =
(
3
2
)
×
(
2
1
)
×
(
3
1
)
×
(
2
2
)
(
2
3
)×(
1
2
)×(
1
3
)×(
2
2
)^4
Choosing 1 red ball and 2 blue balls from one box, and 2 red balls and 1 blue ball from each of the remaining 3 boxes:
(
3
1
)
(
1
3
) ways to choose 1 red ball from one box
(
2
2
)
(
2
2
) ways to choose 2 blue balls from one box
(
3
2
)
(
2
3
) ways to choose 2 red balls from each of the remaining 3 boxes
(
2
1
)
(
1
2
) ways to choose 1 blue ball from each of the remaining 3 boxes
Total ways =
(
3
1
)
×
(
2
2
)
×
(
3
2
)
×
(
2
1
)
(
1
3
)×(
2
2
)×(
2
3
)×(
1
2
)^4
Now, let's calculate each of these expressions.
Let's first calculate the number of ways for each case:
Choosing 1 red ball and 1 blue ball from each of the 4 boxes:
(
3
1
)
(
1
3
) ways to choose 1 red ball from each box
(
2
1
)
(
1
2
) ways to choose 1 blue ball from each box
Total ways =
(
3
1
)
×
(
2
1
)
(
1
3
)×(
1
2
)^4
Choosing 2 red balls and 1 blue ball from one box, and 1 red ball and 2 blue balls from each of the remaining 3 boxes:
(
3
2
)
(
2
3
) ways to choose 2 red balls from one box
(
2
1
)
(
1
2
) ways to choose 1 blue ball from one box
(
3
1
)
(
1
3
) ways to choose 1 red ball from each of the remaining 3 boxes
(
2
2
)
(
2
2
) ways to choose 2 blue balls from each of the remaining 3 boxes
Total ways =
(
3
2
)
×
(
2
1
)
×
(
3
1
)
×
(
2
2
)
(
2
3
)×(
1
2
)×(
1
3
)×(
2
2
)^4
Choosing 1 red ball and 2 blue balls from one box, and 2 red balls and 1 blue ball from each of the remaining 3 boxes:
(
3
1
)
(
1
3
) ways to choose 1 red ball from one box
(
2
2
)
(
2
2
) ways to choose 2 blue balls from one box
(
3
2
)
(
2
3
) ways to choose 2 red balls from each of the remaining 3 boxes
(
2
1
)
(
1
2
) ways to choose 1 blue ball from each of the remaining 3 boxes
Total ways =
(
3
1
)
×
(
2
2
)
×
(
3
2
)
×
(
2
1
)
(
1
3
)×(
2
2
)×(
2
3
)×(
1
2
)^4
Now, let's calculate each of these expressions.
Let's start by calculating the number of ways for each case:
Choosing 1 red ball and 1 blue ball from each of the 4 boxes:
(
3
1
)
(
1
3
) ways to choose 1 red ball from each box
(
2
1
)
(
1
2
) ways to choose 1 blue ball from each box
Total ways =
(
3
1
)
×
(
2
1
)
(
1
3
)×(
1
2
)^4
Choosing 2 red balls and 1 blue ball from one box, and 1 red ball and 2 blue balls from each of the remaining 3 boxes:
(
3
2
)
(
2
3
) ways to choose 2 red balls from one box
(
2
1
)
(
1
2
) ways to choose 1 blue ball from one box
(
3
1
)
(
1
3
) ways to choose 1 red ball from each of the remaining 3 boxes
(
2
2
)
(
2
2
) ways to choose 2 blue balls from each of the remaining 3 boxes
Total ways =
(
3
2
)
×
(
2
1
)
×
(
3
1
)
×
(
2
2
)
(
2
3
)×(
1
2
)×(
1
3
)×(
2
2
)^4
Choosing 1 red ball and 2 blue balls from one box, and 2 red balls and 1 blue ball from each of the remaining 3 boxes:
(
3
1
)
(
1
3
) ways to choose 1 red ball from one box
(
2
2
)
(
2
2
) ways to choose 2 blue balls from one box
(
3
2
)
(
2
3
) ways to choose 2 red balls from each of the remaining 3 boxes
(
2
1
)
(
1
2
) ways to choose 1 blue ball from each of the remaining 3 boxes
Total ways =
(
3
1
)
×
(
2
2
)
×
(
3
2
)
×
(
2
1
)
(
1
3
)×(
2
2
)×(
2
3
)×(
1
2
)^4
Now, let's calculate each of these expressions.
Let's calculate the number of ways for each case:
Choosing 1 red ball and 1 blue ball from each of the 4 boxes:
(
3
1
)
(
1
3
) ways to choose 1 red ball from each box
(
2
1
)
(
1
2
) ways to choose 1 blue ball from each box
Total ways =
(
3
1
)
×
(
2
1
)
(
1
3
)×(
1
2
)^4
Choosing 2 red balls and 1 blue ball from one box, and 1 red ball and 2 blue balls from each of the remaining 3 boxes:
(
3
2
)
(
2
3
) ways to choose 2 red balls from one box
(
2
1
)
(
1
2
) ways to choose 1 blue ball from one box
(
3
1
)
(
1
3
) ways to choose 1 red ball from each of the remaining 3 boxes
(
2
2
)
(
2
2
) ways to choose 2 blue balls from each of the remaining 3 boxes
Total ways =
(
3
2
)
×
(
2
1
)
×
(
3
1
)
×
(
2
2
)
(
2
3
)×(
1
2
)×(
1
3
)×(
2
2
)^4
Choosing 1 red ball and 2 blue balls from one box, and 2 red balls and 1 blue ball from each of the remaining 3 boxes:
(
3
1
)
(
1
3
) ways to choose 1 red ball from one box
(
2
2
)
(
2
2
) ways to choose 2 blue balls from one box
(
3
2
)
(
2
3
) ways to choose 2 red balls from each of the remaining 3 boxes
(
2
1
)
(
1
2
) ways to choose 1 blue ball from each of the remaining 3 boxes
To find the number of ways to choose 10 balls from the 4 boxes such that each box contributes at least one red ball and one blue ball, you can use a combinatorial approach.
Let's break down the problem:
From each box, you need to choose at least one red ball and one blue ball.
You have a total of 4 boxes, and each box contributes 2 types of balls (red and blue).
Now, let's consider the choices:
Choose 1 red ball from each of the 4 boxes:
(
3
1
)
×
(
3
1
)
×
(
3
1
)
×
(
3
1
)
(
1
3
)×(
1
3
)×(
1
3
)×(
1
3
)
Choose 1 blue ball from each of the 4 boxes:
(
2
1
)
×
(
2
1
)
×
(
2
1
)
×
(
2
1
)
(
1
2
)×(
1
2
)×(
1
2
)×(
1
2
)
Now, you need to choose 2 more balls from the remaining pool of 8 (4 boxes * 2 types):
Choose 2 balls from the remaining 8:
(
8
2
)
(
2
8
)
Now, multiply these choices to get the total number of ways:
Total ways
=
(
3
1
)
×
(
3
1
)
×
(
3
1
)
×
(
3
1
)
×
(
2
1
)
×
(
2
1
)
×
(
2
1
)
×
(
2
1
)
×
(
8
2
)
Total ways=(
1
3
)×(
1
3
)×(
1
3
)×(
1
3
)×(
1
2
)×(
1
2
)×(
1
2
)×(
1
2
)×(
2
8
)
Calculating each term individually:
Total ways
=
3
×
3
×
3
×
3
×
2
×
2
×
2
×
2
×
28
Total ways=3×3×3×3×2×2×2×2×28
Total ways
=
1458
×
28
Total ways=1458×28
Total ways
=
40824
Total ways=40824
Therefore, there are 40,824 different ways to choose 10 balls from the 4 boxes such that each box contributes at least one red ball and one blue ball.
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