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19 Sep 2020, 00:54
Kudos
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D
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Difficulty:
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Question Stats:
50% (01:29) correct
50%
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wrong
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In how many ways can five identical balls be distributed in three different boxes?
A)15
B)18
C)20
D)21
E)24
A)15
B)18
C)20
D)21
E)24
19 Sep 2020, 07:12
Kudos
Bookmarks
kawal27
The following is called the SEPARATOR method.
Five identical balls are to be separated into -- at most -- 3 groupings.
Thus, we need five balls and two separators:
OO|OO|O
Every arrangement of the elements above represents one way to distribute the 5 balls among three boxes A, B and C:
OO|OO|O = A gets 2 balls, B gets 2 balls, C gets 1 ball.
OO||OOO = A gets 2 balls, B gets 0 balls, C gets 3 balls.
OOOOO|| = A gets all 5 balls.
And so on.
To count all of the possible distributions, we simply need to count the number of ways to arrange the 7 elements above (the 5 identical balls and the 2 identical separators).
The number of ways to arrange 7 elements = 7!.
But when an arrangement includes identical elements, we must divide by the number of ways each set of identical arrangements can be arranged.
The reason:
When the identical elements swap positions, the arrangement doesn't change, reducing the total number of unique arrangements.
Here, we must divide by 5! to account for the 5 identical balls and by 2! to account for the 2 identical separators:
\(\frac{7!}{5!2!} = 21\)
19 Sep 2020, 04:08
Kudos
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There are 2 important equations for these types of arrangements
If the number of non negative integral solutions for the equation \(x_1 + x_2 +.. + x_n = n\), then the number of ways the distribution can be done is:
(i) \( ^{n + r - 1} C _ {r - 1}\). In this case, value of any variable can be zero.
(ii) \(^{n - 1} C _ {r - 1}\). In this case, minimum value for any variable is 1.
Given that: n = 5 balls, r = 3 boxes and some of the boxes can have zero balls (as nothing is specified that each box should have at least 1 ball).
Therefore the total number of ways = \(^{5 + 3 - 1} C _ {3 - 1}= ^7C_2 = \frac{7 * 6}{2 * 1} = 21\)
Option D
Arun Kumar
If the number of non negative integral solutions for the equation \(x_1 + x_2 +.. + x_n = n\), then the number of ways the distribution can be done is:
(i) \( ^{n + r - 1} C _ {r - 1}\). In this case, value of any variable can be zero.
(ii) \(^{n - 1} C _ {r - 1}\). In this case, minimum value for any variable is 1.
Given that: n = 5 balls, r = 3 boxes and some of the boxes can have zero balls (as nothing is specified that each box should have at least 1 ball).
Therefore the total number of ways = \(^{5 + 3 - 1} C _ {3 - 1}= ^7C_2 = \frac{7 * 6}{2 * 1} = 21\)
Option D
Arun Kumar
