Question 760213: On a shelf there are 4 different mathematics and 3 different additional maths books. In how many ways can this be done if the additional maths books are together. I have tried everything but in vain, the answer should be 724.
here's my workings [M][M][M][M] we can put the three additional maths books between them, and then I get this 3!*5!=720 but the answer is 724...
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! if the math books are together then they can be arranged in 3! ways = 6.
if you put them on the shelf and they have to be together, then they act as if they're 1 so you would get 5! = 120.
6 * 120 = 720
i'm inclined to think that you're correct and the book is wrong.
i tried this with 4 books where 2 of them have to be together.
I labeled the books ABC where C stood for the pair of books that had to be together.
I labeled the 2 books that had to be together as 1, 2.
this is what I got:
With ABC, you get 6 possible permutations.
they are:
ABC
ACB
BAC
BCA
CAB
CBA
when you replace C with 12, you get:
AB12
A12B
BA12
B12A
12AB
12BA
when you permute 12, you get 1,2 and 2,1.
the other half of the permutations would be:
AB21
A21B
BA21
B21A
21AB
21BA
that's a total of 12 permutations which is equal to 3! * 2!.
i don't see any other possible permutation.
for AB, you can have AB12, AB21, A12B, A21B, 12AB, 21AB.
for BA, you can have BA12, BA21, B12A, B21A, 12BA, 21BA.
the books 12 have to be together but the books AB do not.
if it works for 2 and 2, then it should work for 4 and 3.
5! * 3! looks to me like the correct answer.
i have no idea where the additional 4 would come from.
i can't see it in the simple example.
it's unlikely they exist in the more complex example.
i'd stick with 720 even if the book says 724.
the books are not always right.
some books are more right than others.
|
|
|
