Question 11 a and b.

a) 5040ways

b) 576ways

Explanation:

11a) There are 3 different bundles of reading materials each comprising of 4 comic and 3 magazines. The magazines and comics are placed together to form a bundle.

We are to determine the number of ways we can arrange these items in each of the bundle if there are no restrictions.

When there are no restrictions in combining items together, we can place the items in any positions.

Number of comics in one bundle = 4

Number of magazines in one bundle = 3

Total number of books = 4+3 = 7

For 7 reading materials in each bundle, there are 7 places we could pick for the first reading material, 6 for the next reading material, followed by 5, etc. Therefore, we have 7! total ways of arrangement.

7! = 7×6×5×4×3×2×1 = 5040ways

The number of ways we can arrange the items in one bundle if there are no restrictions on individual items to be placed = 5040ways

11b) We are to determine the number of permutations if the order of comic books in each bundle does not change.

For the order of comic books in each bundle not to change and they are arranged in one pile, it means all the comics are placed together.

The number of ways we can arrange 4 comic books placed together = 4!

4! = 4×3×2×1 = 24ways

Let the 4comics represent one entity = 1

The remaining reading materials = 7-4 = 3

The total number of reading material for the 1 entity and the remaining 3 = 3+1 = 4

The number of ways we can arrange these 1 entity and the remaining 3 books = 4!

4! = 4×3×2×1 = 24ways

The number of permutations if the order of comic books in each bundle does not change = 24×24 = 576ways

answer:

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answer:lol Oof

Explanation: