Bunuel wrote:
In how many ways 11 identical marbles be placed in 3 distinct jars such that no jar is empty?
A. 72
B. 54
C. 45
D. 42
E. 36
Are You Up For the Challenge: 700 Level QuestionsLet the number of balls in the jars be a, b, c, where none of the numbers are zero
=> a + b + c = 11
Using formula: Since we need positive integer solutions, we have: (11-1)C(3-1) = 10C2 = 45 ways
Alternative formula: We assign 1 marble to each of a, b and c; now these are already positive. Thus, we have to distribute 8 more marbles among those 3; but now, the distribution allows for 0 marbles assigned => Number of ways = (8+3-1)C(3-1) = 10C2 = 45 ways
[Thus, it is sufficient to simply know the second formula]
Explanation - How do we get these formulae?Let the 11 identical marbles be denoted as M M M M M M M M M M M
We need to put 2 partitions so that we divide the marbles in 3 groups, say for example: M M M | M M M M | M M M M (where | represents the partition)
Let us represent the above as: M M M
P M M M M
P M M M M
This is a 13 letter word with 11 Ms and 2 Ps and if we arrange it, we will get all distributions based on the position of the
Ps
Ex:
M M M
P M M M M
P M M M M => 3, 4, 4 distribution
M M M M M M M M M M
P M
P => 10, 1, 0 distribution; and so on
Note: assigning 0 to a group is allowed
Thus, total arrangements = 13!/(11!2!) = 13C2 = (11+3-1)C(3-1)
Thus, non-negative integer solutions for a+b+c+d+... (r groups) = N is: (N+r-1)C(r-1) where N represents identical objects and r represents distinct groupsLet us now analyse "positive integer solutions": Continuing in the same manner as above:
Let the 11 identical marbles be denoted as M M M M M M M M M M M
We need to put 2 partitions so that we divide the marbles in 3 groups
To ensure that no group ends up getting zero marbles, we assign 1 marble to each group first, leaving us with 8 marbles. We need to divide the marbles in 3 groups, say for example: M M M | M M M M | M (where | represents the partition)
Let us represent the above as: M M M
P M M M M
P M
This is a 10 letter word with 8 Ms and 2 Ps and if we arrange it, we will get all distributions based on the position of the
Ps
Thus, total arrangements = 10!/(8!2!) = 10C2 = (11-1)C(3-1)
Thus, positive integer solutions for a+b+c+d+... (r groups) = N is: (N-1)C(r-1) where N represents identical objects and r represents distinct groups _________________
Sujoy Kumar Datta |
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