Question

(a) twelve skiers are competing in the final round of the olympic freestyle skiing aerial competition. in how many ways can 3 of the skiers finish first, second, and third to win the gold, silver, and bronze medals? 1320
(b) a relay race team has 4 runners who run different parts of the race. there are 16 students on your track team. how many different ways can your coach select students to compete in the race? 43680
(c) the school yearbook has an editor - in - chief and an assistant editor - in - chief. the staff of the yearbook has 15 students. in how many different ways can students be chosen for these 2 positions? 210
just a reminder with how to do the \in how many ways can the letters in a word be rearranged to create different words\.
how many different ways can the letters messages be arranged to create eight - letter \words\?
m - 1; e - 2; s - 3; a - 1; g - 1
\\(\frac{8!}{1!2!3!1!1!}=3,360\\)
(d) how many different ways can the letters htam be arranged to create four - letter \words\? type your answer...
(e) in how many ways can the letters in classes be rearranged to create 7 letter \words\? since the letter s repeats 3 times, some of the permutations will be the same so we will have to eliminate them. type your answer...
(f) how many ways can the letters in mississippi be arranged to create 11 - letter \words\? type your answer...

Explanation:

Part (d)

Step1: Identify the problem type

We need to find the number of permutations of the letters in "HTAM". Since all letters are distinct, we use the permutation formula for \( n \) distinct objects, which is \( n! \) (n factorial). Here, \( n = 4 \) because there are 4 distinct letters (H, T, A, M).

Step2: Calculate the factorial

The formula for \( n! \) is \( n\times(n - 1)\times(n - 2)\times\cdots\times1 \). For \( n = 4 \), we have:
\( 4! = 4\times3\times2\times1 \)
First, calculate \( 4\times3 = 12 \). Then, \( 12\times2 = 24 \). Finally, \( 24\times1 = 24 \).

Step1: Identify the letters and their counts

The word "CLASSES" has 7 letters. Let's list the counts of each letter:

• C: 1
• L: 1
• A: 1
• S: 3
• E: 1
• So, we have a total of \( n = 7 \) letters with a repeated letter (S) that appears \( r = 3 \) times.

Step2: Use the permutation formula for repeated elements

The formula for permutations of \( n \) objects where one object is repeated \( r \) times is \( \frac{n!}{r!} \). Here, \( n = 7 \) and \( r = 3 \).

First, calculate \( 7! \):
\( 7! = 7\times6\times5\times4\times3\times2\times1 = 5040 \)

Then, calculate \( 3! \):
\( 3! = 3\times2\times1 = 6 \)

Now, divide \( 7! \) by \( 3! \):
\( \frac{7!}{3!} = \frac{5040}{6} = 840 \)

Step1: Identify the letters and their counts

The word "MISSISSIPPI" has 11 letters. Let's list the counts of each letter:

• M: 1
• I: 4
• S: 4
• P: 2
• So, we have \( n = 11 \) letters with repeated letters: I (4 times), S (4 times), and P (2 times).

Step2: Use the permutation formula for multiple repeated elements

The formula for permutations of \( n \) objects where there are repeated elements \( r_1, r_2, \cdots, r_k \) is \( \frac{n!}{r_1! \times r_2! \times \cdots \times r_k!} \). Here, \( n = 11 \), \( r_1 = 4 \) (for I), \( r_2 = 4 \) (for S), and \( r_3 = 2 \) (for P).

First, calculate \( 11! \):
\( 11! = 11\times10\times9\times8\times7\times6\times5\times4\times3\times2\times1 = 39916800 \)

Next, calculate \( 4! \) (for I):
\( 4! = 4\times3\times2\times1 = 24 \)

Calculate \( 4! \) (for S):
\( 4! = 24 \) (same as above)

Calculate \( 2! \) (for P):
\( 2! = 2\times1 = 2 \)

Now, multiply the factorials of the repeated elements: \( 4! \times 4! \times 2! = 24\times24\times2 = 1152 \)

Finally, divide \( 11! \) by the product of the factorials of the repeated elements:
\( \frac{11!}{4! \times 4! \times 2!} = \frac{39916800}{1152} = 34650 \)

Answer:

24

Part (e)