Jill has 3 different number cards, 5 different letter cards, and 2 different shape cards. She will randomly choose:

1. One number card
2. Two separate letter cards
3. One shape card

We can solve this problem by considering each step and using combinations where needed.

Step 1: Choosing a number card

Jill has 3 different number cards to choose from. There are $3$ ways to choose one number card.

Step 2: Choosing two letter cards

Jill has 5 different letter cards, and she needs to choose two separate ones. Since the order of the two letter cards doesn't matter, we need to compute the number of combinations of 5 cards taken 2 at a time. The formula for combinations is:

$\binom{n}{r} = \frac{n!}{r!(n-r)!}$

Here, $n = 5$ and $r = 2$. So:

$\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10$

There are 10 ways to choose 2 letter cards.

Step 3: Choosing a shape card

Jill has 2 different shape cards to choose from. There are $2$ ways to choose one shape card.

Step 4: Calculating the total number of ways

To find the total number of ways Jill can choose the four cards, we multiply the number of choices for each step:

$3 \times 10 \times 2 = 60$

Thus, there are 60 different ways to choose the four cards.

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Would you like more details or have any questions?

Here are 5 related questions:

1. How would the solution change if Jill had to choose 3 letter cards instead of 2?
2. What if the order of the letter cards did matter—how would you calculate the total?
3. How would the result change if Jill had 4 shape cards instead of 2?
4. If one more number card was added, how many different ways would there be to choose?
5. What if Jill could select the same letter card twice—how would the total change?

Tip: When order matters in selection, use permutations; otherwise, use combinations.