Catherine rolls a standard 6-sided die six times. If the product of her rolls is 600 then how many different sequences of rolls could there have been? (The order of the rolls matters.)

Answerscorrectly Jul 13, 2024

#1**0 **

If you consider the cases where the order doesn't matter (so 1,2,3,4,5,6 is the same as 6,4,5,3,1,2), there are 4 possibilities. Find those, then use basic combinatorics to find the number of possibilities.

Maxematics Jul 13, 2024

#2**0 **

Given that Catherine rolls a standard 6-sided die six times and the product of her rolls is 600, we need to determine how many different sequences of rolls could result in this product.

Firstly, we factorize 600 into its prime factors:

\[ 600 = 2^3 \times 3 \times 5^2 \]

The numbers on a 6-sided die are 1, 2, 3, 4, 5, and 6. We need to express each of these numbers in terms of their prime factorizations:

- \(1 = 1\)

- \(2 = 2\)

- \(3 = 3\)

- \(4 = 2^2\)

- \(5 = 5\)

- \(6 = 2 \times 3\)

The product of these rolls must equate to \(2^3 \times 3 \times 5^2\). Since each roll can only be 1, 2, 3, 4, 5, or 6, we must distribute the prime factors accordingly among the six rolls.

### Step 1: Distribute the Prime Factors

Each roll will contribute to the overall prime factorization of 600. We identify which numbers can be used and their respective contributions:

- \(2\): contributes \(2\)

- \(3\): contributes \(3\)

- \(4\): contributes \(2^2\)

- \(5\): contributes \(5\)

- \(6\): contributes \(2 \times 3\)

### Step 2: Determine Possible Rolls

To satisfy the factorization \(2^3 \times 3 \times 5^2\) with six rolls:

- We need three 2's.

- We need one 3.

- We need two 5's.

Possible combinations of rolls can be:

- 2 can be contributed by 2, 4 (as \(2^2\)), or 6 (as \(2\)).

- 3 can be contributed by 3 or 6 (as \(3\)).

- 5 can be contributed by 5.

### Step 3: Identify Suitable Rolls

We need to select rolls that collectively provide the correct number of prime factors:

- \(2\) appears in rolls: \(2, 4, 6\).

- \(3\) appears in rolls: \(3, 6\).

- \(5\) appears in rolls: \(5\).

### Step 4: Determine Combinations

Let's break down how we can achieve the required distribution:

- \(2^3\) can be achieved by:

- (2, 2, 2)

- (4, 2)

- (4, 4)

- (2, 6)

- (6, 6)

- \(3\) must be from:

- (3)

- (6)

- \(5^2\) must be from:

- (5, 5)

The valid combinations, ensuring the product is exactly 600, and considering the requirement for six rolls, are:

\[ (2, 2, 2, 3, 5, 5) \]

### Step 5: Counting Permutations

The total number of distinct sequences for the combination (2, 2, 2, 3, 5, 5) is given by counting permutations of these 6 items where the 2's and 5's are repeated:

\[ \frac{6!}{3! \cdot 2!} = \frac{720}{6 \cdot 2} = 60 \]

Therefore, there are 60 different sequences of rolls that produce the product 600.

Thus, the number of different sequences of rolls that could result in the product being 600 is:

\[ \boxed{60} \]

Rangcr897 Jul 19, 2024