Question

A boy has 5 red , 2 yellow and 4 green marbles. In how many ways can the boy arrange the marbles in a line if:
a) Marbles of the same color are indistinguishable?

b) All marbles have different sizes?

Answer 1

_{a).
There are 11 marbles altogether. If they were all different there would be 11! (ie 11x10x9x8x7x6x5x4x3x2x1) ways of arranging them.}

_{The first could be any of 11,}

_{The second could be any of the 10 remaining (one is already placed first)}

_{The third could be any of the 9 remaining (first two are already placed)}

_{The 4th could be any of the 8 remaining (first 3 are already in place)}

_{and so on}

_{Therefore the number of different ways they could be ordered is}

_{11x10x9x8x7x6x5x4x3x2x1 = 11!}

_{However, all the marbles are not different, there are 5xRed, 2xYellow and 4xGreen.}

_{The 11 factorial solution above includes instances where the combinations are the same but the identical Yellow marbles are in the same positions but in a different order.eg}

_YaYb=YbYa

_{The different Yellow could be swopped over with each other without there being any difference in the overall combination. Therefore we must divide 11! by 2(or 2! ie 2x1) to allow for the multiple yellow}

_{In a similar way, we must divide by 5! in respect of the Five identical Red and divide by a further 4! in respect of the identical Green}

_{Therefore the final answer is that there are 11! / 5!2!4! ways}

_{= 6930 combinations}

_b).

_{If marbles have different sizes they can all be distinguished therefore the answer is simply 11!}

_{11x10x9x8x7x6x5x4x3x2x1 = 39916800 different combinations.}