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There are three piles of identical yellow, black and green balls and each pile contains at least 20 balls. The number of ways of selecting 20 balls if the number of black balls to be selected is twice the number of yellow balls is.
- a)6
- b)7
- c)8
- d)9
Correct answer is option 'B'. Can you explain this answer?
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There are three piles of identical yellow, black and green balls and e...
Possible configuration of balls to be selected seven possible cases are there.
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There are three piles of identical yellow, black and green balls and e...
Possible configuration of balls to be selected seven possible cases are there.
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There are three piles of identical yellow, black and green balls and e...
Understanding the Problem
We need to select a total of 20 balls from three piles: yellow, black, and green. The specific condition is that the number of black balls (B) selected must be twice the number of yellow balls (Y). This can be expressed as:
B = 2Y
Let G represent the number of green balls selected. The equation for the total number of balls selected is:
Y + B + G = 20
By substituting B with 2Y, we get:
Y + 2Y + G = 20
This simplifies to:
3Y + G = 20
Finding Possible Combinations
Now, we can express G in terms of Y:
G = 20 - 3Y
Since G must be non-negative, we have:
20 - 3Y ≥ 0
This leads to:
3Y ≤ 20 ---> Y ≤ 6.67
Since Y must be a whole number, the possible values for Y are:
- Y = 0
- Y = 1
- Y = 2
- Y = 3
- Y = 4
- Y = 5
- Y = 6
Calculating Corresponding Values
For each valid Y value, calculate the corresponding G:
- If Y = 0, G = 20 - 3(0) = 20 → B = 0
- If Y = 1, G = 20 - 3(1) = 17 → B = 2
- If Y = 2, G = 20 - 3(2) = 14 → B = 4
- If Y = 3, G = 20 - 3(3) = 11 → B = 6
- If Y = 4, G = 20 - 3(4) = 8 → B = 8
- If Y = 5, G = 20 - 3(5) = 5 → B = 10
- If Y = 6, G = 20 - 3(6) = 2 → B = 12
Counting the Valid Combinations
Each combination of (Y, B, G) is valid as follows:
- (0, 0, 20)
- (1, 2, 17)
- (2, 4, 14)
- (3, 6, 11)
- (4, 8, 8)
- (5, 10, 5)
- (6, 12, 2)
This gives us a total of 7 ways to select 20 balls under the given conditions. Therefore, the correct answer is option B: 7.
We need to select a total of 20 balls from three piles: yellow, black, and green. The specific condition is that the number of black balls (B) selected must be twice the number of yellow balls (Y). This can be expressed as:
B = 2Y
Let G represent the number of green balls selected. The equation for the total number of balls selected is:
Y + B + G = 20
By substituting B with 2Y, we get:
Y + 2Y + G = 20
This simplifies to:
3Y + G = 20
Finding Possible Combinations
Now, we can express G in terms of Y:
G = 20 - 3Y
Since G must be non-negative, we have:
20 - 3Y ≥ 0
This leads to:
3Y ≤ 20 ---> Y ≤ 6.67
Since Y must be a whole number, the possible values for Y are:
- Y = 0
- Y = 1
- Y = 2
- Y = 3
- Y = 4
- Y = 5
- Y = 6
Calculating Corresponding Values
For each valid Y value, calculate the corresponding G:
- If Y = 0, G = 20 - 3(0) = 20 → B = 0
- If Y = 1, G = 20 - 3(1) = 17 → B = 2
- If Y = 2, G = 20 - 3(2) = 14 → B = 4
- If Y = 3, G = 20 - 3(3) = 11 → B = 6
- If Y = 4, G = 20 - 3(4) = 8 → B = 8
- If Y = 5, G = 20 - 3(5) = 5 → B = 10
- If Y = 6, G = 20 - 3(6) = 2 → B = 12
Counting the Valid Combinations
Each combination of (Y, B, G) is valid as follows:
- (0, 0, 20)
- (1, 2, 17)
- (2, 4, 14)
- (3, 6, 11)
- (4, 8, 8)
- (5, 10, 5)
- (6, 12, 2)
This gives us a total of 7 ways to select 20 balls under the given conditions. Therefore, the correct answer is option B: 7.
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