On Halloween, a man presents a child with a bowl containing six different pieces of candy and six different pieces of gum. He tells her that she may have four of each. How many choices does she have?

Answer as a whole number.

Well, the child has quite a few choices, but we can't forget that it's Halloween, so let's calculate it in a spooky way.

We have 6 different pieces of candy and 6 different pieces of gum. Since she can choose 4 of each, it's like she's trick-or-treating at two houses simultaneously.

To calculate the number of choices, we can think of it as 6 choices for the candy house and 6 choices for the gum house. Multiply these two together, and we get a total of 36 choices.

So, the child has 36 choices in total. And hey, that's enough to satisfy any sweet tooth!

To find the number of choices the child has, we need to multiply the number of choices for candy by the number of choices for gum.

The child can choose 4 out of 6 pieces of candy. This can be calculated using the combination formula:

C(6,4) = 6! / (4! * (6-4)!) = 6! / (4! * 2!) = (6 * 5 * 4!)/(4! * 2) = (6 * 5) / 2 = 30 / 2 = 15 choices for candy.

Similarly, the child can choose 4 out of 6 pieces of gum.

C(6,4) = 6! / (4! * (6-4)!) = 6! / (4! * 2!) = (6 * 5 * 4!)/(4! * 2) = (6 * 5) / 2 = 30 / 2 = 15 choices for gum.

To find the total choices, we need to multiply the number of choices for candy by the number of choices for gum:

15 choices for candy * 15 choices for gum = 15 * 15 = 225.

Therefore, the child has 225 choices in total.

To find out how many choices the child has, we need to calculate the combinations of candy and gum she can have.

First, let's calculate the number of ways she can choose four pieces of candy from the six different options. This can be done using the combination formula:

C(n, r) = n! / (r!(n-r)!)

Where n is the total number of options and r is the number of choices.

In this case, the child has 6 options of candy (n) and wants to choose 4 (r). So the calculation would be:

C(6, 4) = 6! / (4!(6-4)!) = 6! / (4!2!)

Let's simplify this fraction:

6! = 6 x 5 x 4!
4! = 4 x 3 x 2 x 1
2! = 2 x 1

Plugging in the values:

C(6, 4) = (6 x 5 x 4 x 3 x 2 x 1) / (4 x 3 x 2 x 1 x 2 x 1)

Now we can simplify and cancel out some terms:

C(6, 4) = (6 x 5) / (2 x 1)
C(6, 4) = 30 / 2
C(6, 4) = 15

So there are 15 different ways the child can choose four pieces of candy from the six options.

Now let's do the same calculation for gum.

C(6, 4) = 15

Since the child can have four of each candy and gum, we need to find the total number of combinations by multiplying the number of choices for candy (15) by the number of choices for gum (15):

Total choices = 15 x 15 = 225

Therefore, the child has 225 different choices in total.

assuming the order of choice does not matter,

6C4 * 6C4
or,
6C2 * 6C2

= 15*15 = 225