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 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 \(n!\). For \(n = 6\) students and \(n=6\) chairs, the formula is \(P(n,n)=\frac{n!}{(n - n)!}=n!\).

$$6! = 6\times5\times4\times3\times2\times1$$

Step2: Calculate the factorial

$$6\times5\times4\times3\times2\times1=720$$

Step3: Permutation formula for \(n = 15\) and \(r=3\)

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

$$P(15,3)=15\times14\times13$$

Step4: Calculate the product

$$15\times14\times13 = 2730$$

Step5: Permutation formula for \(n = 9\) and \(r = 3\)

Using \(P(n,r)=\frac{n!}{(n - r)!}\), with \(n = 9\) and \(r=3\), \(P(9,3)=\frac{9!}{(9 - 3)!}=\frac{9!}{6!}\)

$$P(9,3)=9\times8\times7$$

Step6: Calculate the product

$$9\times8\times7=504$$

Step7: For the number \(3334777\)

The total number of digits \(n = 7\). The digit \(3\) appears \(3\) times, the digit \(4\) appears \(1\) time and the digit \(7\) appears \(3\) times. The formula for permutations of multi - set is \(\frac{n!}{n_1!n_2!n_3!}\), where \(n=n_1 + n_2+n_3\), \(n_1 = 3\) (for \(3\)s), \(n_2=1\) (for \(4\)s) and \(n_3 = 3\) (for \(7\)s)

$$n=7,n_1 = 3,n_2=1,n_3 = 3$$

$$\frac{7!}{3!1!3!}=\frac{7\times6\times5\times4\times3!}{3!\times1\times3!}$$

$$=\frac{7\times6\times5\times4}{3\times2\times1}$$

$$=140$$

Step8: For the bracelet (circular permutation with no distinction for clock - wise and anti - clockwise)

The formula for circular permutations of \(n\) distinct objects is \(\frac{(n - 1)!}{2}\) when considering no distinction between clock - wise and anti - clockwise arrangements. Here \(n=6\)

$$\frac{(6 - 1)!}{2}=\frac{5!}{2}$$

$$=\frac{120}{2}=60$$

Answer:

1. \(720\)
2. \(2730\)
3. \(504\)
4. \(140\)
5. \(60\)