Question

5 Mathematics books, 3 English books and 2 French books are to be arranged in a row on a shelf. Assume that the books of same subject indistinguishable. In how many ways can this be done
(a) without restriction,
(b) with the Mathematics books together,
(c) with the books of same subject grouped together?

Answer 1

(a) The number of permutations without restriction is given by:
\[\frac{10!}{5! \times3! \times2!}=you\ can\ calculate\]

Answer 2

@vernonwan Do you understand?

Answer 3

why not the answer of (a) to be 10!

Answer 4

The reason being that 10! would be correct only if all of the books were different. However in this question, all 5of the mathematics books are exactly the same, all 3 of the English books are exactly the same and both of the French books are exactly the same. Therefore 10! must be divided by the number of permutations of the books in each subject. This last step eliminates the permutations among the 10! which are exactly the same.

Answer 5

(b) If the 5 mathematics books must be together, they can be considered as one book (five times as wide as each of the mathematics books).
Using the same reasoning as for (a) the number of permutations is given by:
\[\frac{6!}{3! \times2!}=you\ can\ calculate\]

Answer 6

(c) If books of the same subject are grouped together they can be considered as just 3 different books. Therefore the number of permutations is 3!

Answer 7

4 different letters are selected from the word 'LOGARITHM'
Use two methods to find the number of ways of selecting at least one vowel.
thanks

Answer 8

@vernonwan Sorry but I must log out now. Please post your last question as a new question.

Answer 9

thanks

Answer 10

You're welcome :)