Question

iii. what can i do with what i know: solve the following problems. write your solutions and answers on a separate sheet of paper. (3 points each) 1. how many ways can 6 students be seated in a row of 6 chairs for pictorial? 2. if fifteen young ladies compete for the first three most beautiful ladies in a pageant, in how many ways can the winners be selected assuming that there are no ties? 3. nine students are to be seated in a row of chairs. how many arrangements are possible if 3 students insist on sitting beside each other? 4. how many different numbers can be formed using all the digits of 3 3 3 4 7 7 7? 5. in how many ways can 9 different colored beads be arranged on a bracelet?

Explanation:

Step1: Permutation formula

The number of permutations of \(n\) distinct objects taken \(n\) at a time is \(P(n,n)=\frac{n!}{(n - n)!}=n!\)
For \(n = 6\) (6 students and 6 chairs), the number of arrangements is \(6!\)

Step2: Calculate \(6!\)

\(6!=6\times5\times4\times3\times2\times1 = 720\)

Step1: Permutation formula for \(n\) objects taken \(r\) at a time

The number of permutations of \(n\) distinct objects taken \(r\) at a time is \(P(n,r)=\frac{n!}{(n - r)!}\)
Here \(n = 15\) (15 young ladies) and \(r=3\) (selecting 3 winners)
\(P(15,3)=\frac{15!}{(15 - 3)!}=\frac{15!}{12!}\)

Step2: Simplify the expression

\(\frac{15!}{12!}=15\times14\times13\)
\(15\times14\times13=(10 + 5)\times14\times13=(140+70)\times13 = 210\times13=2730\)

Step1: Permutation formula for circular permutations

The number of circular permutations of \(n\) distinct objects is \((n-1)!\)
Here \(n = 9\) (9 students), but we are arranging 3 students in a row (linear permutation).
The number of ways to choose 3 students out of 9 is \(P(9,3)=\frac{9!}{(9 - 3)!}=\frac{9!}{6!}\)

Step2: Simplify the expression

\(\frac{9!}{6!}=9\times8\times7=504\)

Step1: Analyze the digits of the number

The number \(3334777\) has \(7\) digits. The digit \(3\) appears \(3\) times, the digit \(4\) appears \(1\) time and the digit \(7\) appears \(3\) times.
The number of permutations of a multi - set with \(n\) objects, where there are \(n_1\) of one kind, \(n_2\) of another kind,\(\cdots\), \(n_k\) of the \(k\)th kind is \(\frac{n!}{n_1!n_2!\cdots n_k!}\)
Here \(n=7\), \(n_1 = 3\) (for digit \(3\)), \(n_2=1\) (for digit \(4\)) and \(n_3 = 3\) (for digit \(7\))
\(\frac{7!}{3!1!3!}\)

Step2: Calculate the factorials

\(7! = 5040\), \(3! = 6\), \(1! = 1\)
\(\frac{5040}{6\times1\times6}=\frac{5040}{36}=140\)

Answer:

720