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In how many ways can 4 red, 3 yellow and 2 green discs be arranged in a row, if the discs of the same colour are indistinguishable?
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The correct answer is:
1260
Total number of dises are $4+3+2=9$.
Out of 9 dises, 4 are of the first kind (red) 3 are of the second kind (yellow) and 2 are of the third kind (green).
Therefore, the number of arrangements
$=\frac{9 !}{4 ! 3 ! 2 !}=1260$
Out of 9 dises, 4 are of the first kind (red) 3 are of the second kind (yellow) and 2 are of the third kind (green).
Therefore, the number of arrangements
$=\frac{9 !}{4 ! 3 ! 2 !}=1260$
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