18. crafting jaclyn bought some decorative letters for a scrapbook project. if she randomly selected a…

Explanation:

Step1: Identify the letters in "photography"

First, we list the letters in the word "photography" and count the frequency of each letter:

• p: 2
• h: 2
• o: 1
• t: 1
• g: 2
• r: 1
• a: 1
• y: 1

The total number of letters in "photography" is ( n = 12 ) (since ( 2 + 2+1 + 1+2 + 1+1 + 1=12 )).

Step2: Calculate the number of permutations of the letters

The formula for the number of permutations of a multiset (when there are repeated elements) is ( \frac{n!}{n_1!n_2!\cdots n_k!} ), where ( n ) is the total number of elements, and ( n_1, n_2, \cdots, n_k ) are the frequencies of the repeated elements.

For the word "photography", the number of distinct permutations of its letters is ( \frac{12!}{2! \times 2! \times 2!} ) (because the letters p, h, and g are each repeated 2 times).

But when we randomly select a permutation of these letters, the number of favorable outcomes (the permutation that forms "photography") is 1, and the total number of possible permutations of these 12 letters (with repeated elements) is ( \frac{12!}{2! \times 2! \times 2!} ).

Step3: Calculate the probability

The probability ( P ) is given by the number of favorable outcomes divided by the number of total outcomes.

The number of favorable outcomes (forming "photography") is 1.

The total number of possible permutations of the 12 letters (with p repeated 2 times, h repeated 2 times, and g repeated 2 times) is ( \frac{12!}{2! \times 2! \times 2!} ).

So the probability ( P=\frac{1}{\frac{12!}{2! \times 2! \times 2!}}=\frac{2! \times 2! \times 2!}{12!} )

We know that ( n!=n\times(n - 1)\times\cdots\times1 ), so:

( 12! = 12\times11\times10\times9\times8\times7\times6\times5\times4\times3\times2\times1=479001600 )

( 2! = 2\times1 = 2 ), so ( 2! \times 2! \times 2! = 2\times2\times2=8 )

Then ( P=\frac{8}{479001600}=\frac{1}{59875200} )

Answer:

( \frac{1}{59875200} )