A boy has 4 red , 6 yellow and 3 green marbles. In how many ways can the boy arrange the marbles in a line if:
a) Marbles of the same color are indistinguishable?
Maths problems?
2016-09-19 7:16 am
回答 (3)
2016-09-19 7:56 am
(4+6+3)!/[4!*6!*3!]
2016-09-19 8:03 am
(13)!/[4!*6!*3!] = ...calculator
2016-09-19 7:55 am
Let's first consider the number of permutations if all 13 marbles were distinguishable. That would just be 13! = 13 x 12 x 11 x ... x 3 x 2 x 1 = 6,227,020,800
Now for every permutation, there will be 4! = 24 arrangements that are identical except for the fact that the red balls are rearranged. So we need to divide by 24 to account for the fact that they are indistinguishable from each other.
Likewise for the 6 yellow balls there are 6! = 720 identical arrangements for each given permutation, so divide by 720 to account for that.
And finally for the 3 green balls, we divide by 3! = 6 to account for those identical arrangements.
The final formula is:
13! / (4! 6! 3!)
= 6227020800 / (24 * 720 * 6)
= 60,060
Answer:
60,060 ways
Now for every permutation, there will be 4! = 24 arrangements that are identical except for the fact that the red balls are rearranged. So we need to divide by 24 to account for the fact that they are indistinguishable from each other.
Likewise for the 6 yellow balls there are 6! = 720 identical arrangements for each given permutation, so divide by 720 to account for that.
And finally for the 3 green balls, we divide by 3! = 6 to account for those identical arrangements.
The final formula is:
13! / (4! 6! 3!)
= 6227020800 / (24 * 720 * 6)
= 60,060
Answer:
60,060 ways
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