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# Combination of marbles

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Tom has 6 red marbles, 4 blue marbles, and 3 green marbles. How many distinct ways can he line them up in a row?

A) 60060

B) 288288

C) 3459456

D) 10378368

E) 6227020800

Aug 31, 2019

#1
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$$\text{first place the red marbles, then the blue, the green go into the remaining slots}\\ N=\dbinom{13}{6}\dbinom{7}{4} = 60060$$

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Sep 1, 2019
#2
+4

Tom has 6 red marbles, 4 blue marbles, and 3 green marbles.

How many distinct ways can he line them up in a row?

$$\begin{array}{|rcll|} \hline && \mathbf{\dfrac{(6+4+3)!}{6!4!3!}} \\\\ &=& \dfrac{13!}{6!4!3!} \\\\ &=& \dfrac{6!\cdot 7\cdot 8\cdot 9\cdot 10\cdot 11\cdot 12\cdot 13}{6!4!3!} \\\\ &=& \dfrac{7\cdot 8\cdot 9\cdot 10\cdot 11\cdot 12\cdot 13}{4!3!} \\\\ &=& \dfrac{7\cdot 8\cdot 9\cdot 10\cdot 11\cdot 12\cdot 13}{1\cdot 2\cdot 3\cdot 4\cdot 1\cdot 2 \cdot 3} \\\\ &=& 7\cdot 3\cdot 5\cdot 11\cdot 4\cdot 13 \\ &=& \mathbf{60060} \\ \hline \end{array}$$ Sep 1, 2019
#3
+3

I'll give an explanation on why it is.

The red marbles, blue marbles, and green marbles that Tom has can be arranged as RRRRRRBBBBGGG, where R denotes the red marbles, B denotes the blue marbles, and G denotes the green marbles.

Tom has, therefore, 6+4+3=13 marbles, so we would assume that if each ball was its own different color, then there would be 13!(Thirteen Factorial) ways to arrange the balls. However, there are six(6) red balls, four(4) blue balls, and three(3) green balls, so these are duplicates of the three colors(red, blue, and green). Thus, the number of ways to arrange the thirteen balls are (13!) / (6! * 4! * 3!), $$\frac{13!}{6!*4!*3!}$$which is $$4*7*3*5*11*13=\boxed{60060}$$ distinct ways that Tom can line the thirteen balls up.

Sep 1, 2019