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Bunuel
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Bunuel
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Bunuel
If the question had been about unique Bs & Gs, then the answer should have been 7*5!*6!, correct?

No, in this case the answer would be 7!*5!.

Glue b's together, we'll get 7 units:
{bbbbb}{g}{g}{g}{g}{g}{g}

# of arrangements = 7!. 5 g's within their unit can be arranged in 5! ways. So, 7!*5!.
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In how many ways 5 identical blue marbles and 6 identical green marbles can be arranged in a row, so that all the blue marbles are together?

A. 120
B. 30
C. 24
D. 11
E. 7

Let, all the blue marbles are one, so the total element is 6+1=7, Green marbles are 6.

Here the formula is \(=\frac{7!}{6!}=7\)

The answer is E
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In how many ways 5 identical blue marbles and 6 identical green marbles can be arranged in a row, so that all the blue marbles are together?

A. 120
B. 30
C. 24
D. 11
E. 7

Think of the 5 identical blue marbles as one unit, so it gives us 1+6 = 7 units, these can be arranged in 7! ways BUT we have 6 identical units in these 7 units. We know that in combinatorics we have to divide by the number of identical units so the answer = 7!/6!= 7
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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I like the solution - it’s helpful.
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I like the solution - it’s helpful.
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