SOLUTION: I need help. Four different math books, two different English books, three different history books are arranged in a row. Find the probability that a) the books are grouped togethe
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Question 1095708: I need help. Four different math books, two different English books, three different history books are arranged in a row. Find the probability that a) the books are grouped together according to their subject area.
b) the English books are not grouped together.
Thank you.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
you have 3 groups.
keeping the groups together, you can arrange them on the shelf 3! ways.
let the groups be a,b,c
you have:
abc
acb
bac
bca
cab
cba
within each group, however, they can be arranged as follows:
group a has 4 books which can be arranged in 4! ways.
group b has 2 books which can be arranged in 2! ways.
group c has 3 books which can be arranged in 3! ways.
the total possible arrangements are 3! * 4! * 2! * 3!.
3! = 6
4! = 24
2! = 2
3! = 6
you get 6 * 24 * 2 * 6 which is equal to 1728.
the books can be arranged on the shelf 1728 ways.
to see how this works, lets take 2 different groups with 2 different books in the first and 2 different books in the second group.
let the groups be a,b
let the individual books in group a be a1,a2.
let the individual books in group b be b1,b1.
the 2 groups a and b can be arranged in 2! ways.
that would be a,b and b,a
within group a, you can arrange the books in 2! ways.
that would be a1,a2 and a2,a1
within group b, you can arrange the books in 2! ways.
that would be b1,b2 and b2,b1
the possible shelf arrangements are:
a1,a2,b1,b2
a1,a2,b2,b1
a2,a1,b1,b2
a2,a1,b2,b1
b1,b2,a1,a2
b1,b2,a2,a1
b2,b1,a1,a2
b2,b1,a2,a1
that's a total of 8 arrangements
that's equal to 2! for the arrangements of the groups * 2! for the arrangements of the 2 books within the a group * 2! for the arrangement of the 2 books within the b group.
2! * 2! * 2! = 2 * 2 * 2 which is equal to 8.
the same concept applies to your original problem.
3! for the 3 groups * 4! for the arrangement of the 4 math books within the first group * 2! for the arrangement of the 2 math books within the second group * 3! for the 3 history books within the third group.
3! * 4! * 2! * 3! = 6 * 24 * 2 * 6 = 1728
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