Seven blue and four red balls are to be arranged in order. How many ways can this be done if
(1) The blue balls are distinguishable (e.g. numbered) as are the red balls.
(2) Blue balls are distinguishable, but the red balls are identical.
(3) The balls of each color are indistinguishable.
+6248 \(\text{I don't think you mean in order. I think you mean how many distinct arrangements are there}\\ \text{1) there are 11 total distinct balls. There are 11! = 39916800 arrangements}\\ \text{2) there are 4 identical balls so we have 4! identical arrangements. This gives a total of $\dfrac{11!}{4!} = 1663200$ arrangements}\\ \text{3) $\dfrac{11!}{7!4!} = 330$ arrangements}\)
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