Review of the Previous Discussions

There are three men and 3 women There are three men and 3 women. In how many ways can they be seated in a row of 6 chairs if men and women alternate? there are two cases: M-W-M-W-M-W or W-M-W-M-W-M in each case, there are: 3 x 3 x 2 x 2 x 1 x 1 ways = 36 total for the 2 cases = 2 x 36 = 72 different arrangements possible

2. In how many ways can 4 men and 3 women be seated in a row of 7 chairs if men and women must alternate? The only possible case is M-W-M-W-M-W-M Therefore, there will be 4x3x3x2x2x1x1 = 144 ways 3. In number 2, in how may ways can they be arranged if men and women alternate and a woman occupies the leftmost chair? This is NOT possible because the setup would have been W-M-W-M-W-M-M. 0 ways.

In how many ways can the letters of the word “kababalaghan” be arranged in a row? There are 12 letters----5 of which are A’s and 2 are B’s 12! ÷ (5! x 2!) = 1,995,840 ways 5. There are 5 identical red marbles and 3 identical blue marbles. In how many ways can they be arranged in: a. a row? 8! ÷ (5! x 3!) = 56

There are 5 identical red marbles and 3 identical blue marbles There are 5 identical red marbles and 3 identical blue marbles. In how many ways can they be arranged in: b. a row if the red marbles were labelled 1 to 5? The red marbles now become distinct from one another since each one as a different number written on it. Thus, only the blue ones are identical. 8! ÷ 3! = 6720 ways

5. There are 5 identical red marbles and 3 identical blue marbles 5. There are 5 identical red marbles and 3 identical blue marbles. In how many ways can they be arranged in: c. a row if the marbles were labelled from 1 to 8? All the marbles becomes distinct because each one has a unique number written on it. Thus, there would be 8! = 40320

In how many ways can nine people be arranged in a circle? (9-1)! = 8! = 40320 In how many ways can you arrange 8 people in a row if there are 9 people to choose from? 9P8 = 362,880

8. In how many ways can 3 freshmen and 6 sophomores be arranged in a row if: a. The freshmen are to sit together? The 3 freshmen will be counted as 1. 1 + 6 sophomores = 7 7! x 3! = 30,240 b. The sophomores are to sit together? The 6 sophomores will be counted as 1. 1 + 3 freshmen = 4. 4! x 6! = 17,280 c. The freshmen sit together and so do the sophomores? The freshmen will be counted as 1, the sophomores as well. 1 + 1 = 2. 2! x 3! x 6! =8,640

8. In how many ways can 3 freshmen and 6 sophomores be arranged in a row if: d. The three freshmen are not allowed to sit together? To solve problems like this, solve first for the number of arrangements possible for the people (or objects) where there are no restrictions.  9 people in a row  9! = 362880 ways Then solve for the number of ways where the three freshmen are seated together  7! X 3! = 30,240 There are 362,880 ways of arranging 9 people in a row, 30, 240 of which are arrangements where the 3 freshmen are seated together. That would mean all the other arrangements are ones where the 3 freshmen are not together! Thus, subtract the two, 362,880 – 30,240 = 332,640 ways!

Checking of Homework Each item solved is worth 2 points. 1 point for solution, 1 point for the answer. TOTAL POINTS: 22

1. There are 10 participants in a Math contest 1. There are 10 participants in a Math contest. Each one has an equal chance of winning it. How many possible outcomes are there (they will be ranked from top to bottom)? 2. In a reality show, only the Top 6 in rankings of the 15 participants will make it to the finals. In how many ways can the 6 finalists be chosen? 3. During a workshop of student leaders, participants were grouped by 8’s and asked to seat in circles. In how many ways can the members of one group be seated in the formation?

A box contains 8 white and 2 blue handkerchiefs A box contains 8 white and 2 blue handkerchiefs. In how many ways can they be arranged in a shelf if they have to put on top of one another? In number 4, if the white handkerchiefs were labelled 1 to 8 respectively using a tape, in how many ways can they be arranged? If in number 4, 4 white hankies were labelled with the letter A and the other 4 were labelled with the letter B, and the blue ones were labelled with the letter C, how many arrangements are possible?

There are 4 men and 4 women in a room There are 4 men and 4 women in a room. In how many ways can they be arranged in a line if a. The men and women must alternate? b. The women are to sit together? c. The men are to sit together? d. The men are to sit together, and so are the women? e. The 4 men are not allowed to sit together?

1. There are 10 participants in a Math contest 1. There are 10 participants in a Math contest. Each one has an equal chance of winning it. How many possible outcomes are there (they will be ranked from top to bottom)? 10! = 3,628, 800 ways 2. In a reality show, only the Top 6 in rankings of the 15 participants will make it to the finals. In how many ways can the 6 finalists be chosen? 15P6 = 3,603,600 3. During a workshop of student leaders, participants were grouped by 8’s and asked to seat in circles. In how many ways can the members of one group be seated in the formation? (8-1)! = 7! = 5040

A box contains 8 white and 2 blue handkerchiefs A box contains 8 white and 2 blue handkerchiefs. In how many ways can they be arranged in a shelf if they have to put on top of one another? 10! / (8! X 2!) = 45 5. In number 4, if the white handkerchiefs were labelled 1 to 8 respectively using a tape, in how many ways can they be arranged? 10! / 2! = 1,814,400 6. If in number 4, 4 white hankies were labelled with the letter A and the other 4 were labelled with the letter B, and the blue ones were labelled with the letter C, how many arrangements are possible? 10! / (4! X 4! X 2!) = 3,150

There are 4 men and 4 women in a room There are 4 men and 4 women in a room. In how many ways can they be arranged in a line if a. The men and women must alternate? two cases: M-W-M-W-M-W-M-W or W-M-W-M-W-M-W-M Each case has 4x4x3x3x2x2x1x1 = 576 Total arrangements = 2 x 576 = 1152 b. The women are to sit together? The women will be counted as one 4 men + 1 = 5 5! X 4! = 2880

c. The men are to sit together? 5! X 4! = 2880 7. There are 4 men and 4 women in a room. In how many ways can they be arranged in a line if c. The men are to sit together? 5! X 4! = 2880 d. The men are to sit together, and so are the women? The men will be counted as one, so are the women. 2! X 4! X 4! = 1152 e. The 4 men are not allowed to sit together? # of arrangements possible – #of arrangements where all men are together = 8! – 2880 = 40320 – 2880 = 37440

Seatwork ¼ sheet of paper. Five minutes per slide. Show solutions. Each item solved is worth 2 points

A box contains 5 white and 4 blue handkerchiefs A box contains 5 white and 4 blue handkerchiefs. In how many ways can they be arranged in a shelf if you put handkerchiefs of the same color on top the other bunch of handkerchiefs of the same color? 2 cases. (5!/5!) x (4! x 4!) x 2 cases = 2 ways In number 1, how many row arrangements are possible for the 5 white and 4 blue handkerchiefs? 9! ÷ (5! X 4!) = 126 ways If the white handkerchiefs were each labelled respectively with A to E, while the blue ones were labelled with the number 1, how many row arrangements are possible? 9! ÷ 4! = 15,120 ways

There are 4 freshmen and 3 sophomores There are 4 freshmen and 3 sophomores. In how many ways can you arranged them: a) in a row if all the freshmen sit together? 4! x 4! = 576 ways b) in a row if students from the same level must sit together? 2! x 4! x 3! = 288 ways c) in a circle if students from the same level must sit together? (2 – 1)! x 4! x 3! = 144 ways

There are 4 freshmen and 3 sophomores There are 4 freshmen and 3 sophomores. In how many ways can you arranged them: d) in a row if you alternate freshmen and sophomores? only one case--- F – S – F –S – F – S – F 4 x 3 x 3 x 2 x 2 x 1 x 1 = 144 ways e) in a row if the freshmen are not allowed to sit together? 7! – (4! x 4!) = 5040 – 4,464 ways f) in a row if a freshman must occupy the leftmost seat while a sophomore must occupy the rightmost seat? -start assigning in the two end seats... -4 choices for the leftmost seat, 3 choices for the rightmost seat. -the middle chairs can be occupied by any of the remaining students = 4 x 5 x 4 x 3 x 2 x 1 x 3 = 1,440 ways