+12 To solve this problem, we can use the principle of inclusion-exclusion.
First, we find the total number of ways to distribute the balls without any restrictions. This is equivalent to finding the number of ways to put 10 indistinguishable balls into 8 distinguishable boxes, which is given by the formula:
$${10 + 8 - 1 \choose 8 - 1} = {17 \choose 7} = 19,448$$
Next, we find the number of ways to distribute the balls such that no box is empty. This is equivalent to finding the number of ways to put 10 indistinguishable balls into 8 distinguishable boxes such that each box has at least one ball. We can think of this as a stars and bars problem with the additional constraint that each box must have at least one star (ball). Using the stars and bars formula, we get:
$${10 - 1 \choose 8 - 1} = {9 \choose 7} = 36$$
However, we are interested in the number of ways such that at least one box is empty. To find this, we use the principle of inclusion-exclusion. Let A be the event that box 1 is empty, B be the event that box 2 is empty, and so on up to H for box 8. Then the number of ways to distribute the balls such that at least one box is empty is given by:
$$\sum_{i=1}^{8} (-1)^{i-1} \binom{8}{i} {10 - i - 1 \choose 8 - 1}$$
where the $(-1)^{i-1}$ term comes from the alternating signs in the inclusion-exclusion formula.
Using this formula, we get:
$$\sum_{i=1}^{8} (-1)^{i-1} \binom{8}{i} {10 - i - 1 \choose 8 - 1} = 19,448 - 36\binom{8}{1} + \binom{8}{2} {1 \choose 7} - \binom{8}{3} {2 \choose 7} + \cdots + (-1)^{8-1} \binom{8}{8} {2 \choose 7}$$
Simplifying this expression gives:
$$19,448 - 8\times36 + 28\times0 - 56\times0 + 70\times0 - 56\times0 + 28\times0 - 8\times0 + 1\times0 = 19,064$$
Therefore, the number of ways to distribute 10 indistinguishable balls among 8 distinguishable boxes, if at least one of the boxes must be empty, is 19,064. Subaru Net
+20 
+11 Hello gm dear,
very nice question , well now we will see how it can be solved
There are two cases to consider: one where the two same-colored beads are next to each other, and one where they are not.
Case 1: Two same-colored beads are next to each other
In this case, we can treat the two same-colored beads as a single "unit". We then have five units to arrange: the "unit" of the two same-colored beads and the four units of the other color. The number of ways to arrange five units is 5! = 120. However, we must divide by 2! to correct for overcounting the arrangements of the two same-colored beads within the "unit". Therefore, there are 120/2! = 60 arrangements in this case.
Case 2: Two same-colored beads are not next to each other
In this case, we can treat the two same-colored beads as two separate units. We then have six units to arrange: the two units of the same-colored beads and the four units of the other color. The number of ways to arrange six units is 6! = 720. However, we must divide by 2! to correct for overcounting the arrangements of the two same-colored beads within their units, and we must also divide by 2 to correct for overcounting the arrangements that are reflections of each other. Therefore, there are 720/(2! * 2) = 180 arrangements in this case.
Overall, the total number of different ways that Joanna can assemble her bracelet is the sum of the two cases, which is 60 + 180 = 240.
hope so it will helps or you can see at BenefitsCal App
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