Question 1194811: 16 books are to be lined up on the shelf. If 6 of the books are identical math books and 7 are identical Science books and 3 are identical English books, how many ways can they be lined up?
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
If all 16 books were distinguishable from one another, then we'd have 16! = 20,922,789,888,000 possible permutations. The exclamation mark indicates a factorial.
16! = 16*15*14*...*3*2*1 = 20,922,789,888,000
This massive number has 14 digits and the number is a bit under 21 trillion.
That would be the answer if we could tell every book apart. However, 6 of the books are identical math books which we cannot tell apart. This means we have to divide by 6! = 6*5*4*3*2*1 = 720 to correct for the erroneous overcounting.
The same goes for the 7 identical science books and 3 identical English books.
7! = 7*6*5*4*3*2*1 = 5040
3! = 3*2*1 = 6
Overall we have to divide by 6!*7!*3! = 720*5040*6 = 21,772,800
We get to
(20,922,789,888,000)/(21,772,800) = 960,960
Answer: 960,960
It's not a typo that the "960" is written twice.
Delete the comma if necessary.
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