Seven year-old Sarah has nine crayons; three blue, one red, two green and three yellow.

a) In how many ways can she line up the crayons on her desk?
b) She is to pick up some of the crayons. How many different choices of some crayons could she make?

a) You should have a formula or procedure for that type of question

You are arranging n elements of which m are the same of one kine, k are the same of another kind, p are the same of another kind etc
number of ways = n!/(m!k!p!..)

so here number of ways to arrange the crayons
= 9!/(3!2!3!) = 5040

b) She can take the 3 blue crayons in 4 ways:
take none at all, take 1, take 2, take 3 or take 4
for each of those she can take the red crayons in 2 ways, the greens in 3 ways and the yellows in 4 ways
number of ways = 4x2x3x4 = 96 ways
BUT, that would include the single case of no crayons of any of the colours.
So final number of ways = 96-1 =95

a) Well, if Sarah wants to line up her crayons on her desk, we need to calculate the total number of permutations. Since she has nine crayons, we can calculate this using factorial math. So, it would be 9 factorial, which is 9!. But don't worry, you don't have to solve that right now! The answer is a really big number, somewhere around 362,880. So, Sarah has a whopping 362,880 ways to line up her crayons on her desk. That's cray-on-tastic!

b) Now, let's talk about Sarah's choices for picking some crayons. Since she can pick any combination, we need to calculate the total number of subsets. Luckily, there's a shortcut for this! It's called the power set, and it's calculated by raising 2 to the power of the number of crayons she has. So, 2 to the power of 9 is... *drumroll*... 512! That's right, Sarah has 512 different choices of some crayons she can make. She better start coloring outside the lines with all those options!

a) To find the number of ways Sarah can line up the crayons on her desk, we can use the concept of permutations.

Since Sarah has a total of 9 crayons, we have 9 options for the first position, 8 options for the second position, 7 options for the third position, and so on, until we reach 1 option for the last position.

Therefore, the number of ways Sarah can line up the crayons on her desk is 9! (read as "9 factorial"), which equals 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1. Evaluating this expression gives us the answer:

9! = 362,880.

So, Sarah can line up the crayons on her desk in 362,880 different ways.

b) To find the number of different choices Sarah can make when picking some of the crayons, we need to consider the possible combinations.

Since Sarah has 9 crayons in total, she has the option to choose any number of crayons ranging from 0 (choosing none) up to 9 (choosing all).

To calculate the total number of different choices, we can find the sum of the possible combinations for each number of crayons chosen.

This can be expressed as:

C(0,9) + C(1,9) + C(2,9) + C(3,9) + C(4,9) + C(5,9) + C(6,9) + C(7,9) + C(8,9) + C(9,9),

where C(n,k) represents the number of combinations of choosing k items from a total of n items.

Evaluating the expression gives us:

1 + 9 + 36 + 84 + 126 + 126 + 84 + 36 + 9 + 1 = 512.

So, Sarah can make a total of 512 different choices when picking some of the crayons.

To find the answers to these questions, we can use the concept of permutation and combination.

a) In how many ways can she line up the crayons on her desk?

Since Sarah has nine crayons, we need to find the number of ways she can arrange these nine crayons in a line. To do this, we can use the formula for permutations of objects.

To find the number of permutations, we need to multiply the number of choices for each position. In this case, the first position can be filled with any of the nine crayons, the second position can be filled with any of the remaining eight crayons, and so on.

So, the number of ways to line up the crayons on her desk can be calculated as:

9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 362,880

Therefore, there are 362,880 ways for Sarah to line up the crayons on her desk.

b) She is to pick up some of the crayons. How many different choices of some crayons could she make?

In this case, we need to find the number of combinations that can be formed by selecting some crayons from the total of nine crayons. To do this, we can use the formula for combinations of objects.

The formula for combinations is given by: nCr = n! / (r!(n-r)!), where n is the total number of objects and r is the number of objects being selected.

In Sarah's case, she has nine crayons, and she can choose any number of crayons from zero up to nine. We need to find all possible combinations for each possible value of r.

So, by applying the combination formula for each value of r, we can find the total number of different choices of some crayons Sarah could make:

nCr = 9! / (r!(9-r)!)

0C9 = 9! / (9!(9-9)!) = 1
1C9 = 9! / (9!(9-1)!) = 9
2C9 = 9! / (9!(9-2)!) = 36
3C9 = 9! / (9!(9-3)!) = 84
4C9 = 9! / (9!(9-4)!) = 126
5C9 = 9! / (9!(9-5)!) = 126
6C9 = 9! / (9!(9-6)!) = 84
7C9 = 9! / (9!(9-7)!) = 36
8C9 = 9! / (9!(9-8)!) = 9
9C9 = 9! / (9!(9-9)!) = 1

By summing up the combinations for each value of r, we get:

1 + 9 + 36 + 84 + 126 + 126 + 84 + 36 + 9 + 1 = 517

Therefore, there are 517 different choices of some crayons that Sarah could make.