In how many ways can 8 people be seated in a row if a.) there are no restrictions on the seating arrangement(answer is 8!=40,320) , b.) persons A and B must sit next to each other? (Answer: 10,080) , c.) there are 4 men and 4 women and no 2 men or 2 women can sit next to each other? (Answer: 1,152) , d.) there are 5 men and they must sit next to one another? (Answer: 2,880) , e.) there are 4 married couples and each couple must sit together? (Answer: 384). I need to see the steps to solve questions 10b to 10e. I quickly looked through the corresponding chapter on combinatorial analysis and I went over topics such as the multinomial theorem, the basic principle of counting, the binomial theorem, and representing the number of possible combinations of n objects taken r at a time as "n choose r" .

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In how many ways can 8 people be seated in a row if a.) there are no restrictions on the seating arrangement(answer is 8!=40,320) , b.) persons A and B must sit next to each other? (Answer: 10,080) , c.) there are 4 men and 4 women and no 2 men or 2 women can sit next to each other? (Answer: 1,152) , d.) there are 5 men and they must sit next to one another? (Answer: 2,880) , e.) there are 4 married couples and each couple must sit together? (Answer: 384). I need to see the steps to solve questions 10b to 10e. I quickly looked through the corresponding chapter on combinatorial analysis and I went over topics such as the multinomial theorem, the basic principle of counting, the binomial theorem, and representing the number of possible combinations of n objects taken r at a time as &quot;n choose r&quot; .

Dawn Neal · Accepted Answer

Step 1  Given that there are 8 people sitting in a row.  The number of people, n=8.  Step 2  (a) To find the number of possible ways in which 8 people can be seated in a row, when there are no restrictions.  The number of ways in which n people can be seated in a row, when there are no restrictions =n!  Therefore,  The number of ways         =n!                                                   =8!                                                   8×7×6×5×4×3×2×1                                                   =40,320  Step 3  (b) To find the number of possible ways in which 8 people can be seated in a row, when A and B must sit next to each other.  Consider A and B be a single block and then proceed. Now, there are 6 people and one block consisting of A and B. Observe that there are 7 positions to permute with.  The number of ways in which n people can be seated in a row, when there are no restrictions =n!  Therefore,  The number of ways         =n!                                                   =n!                                                   7×6×5×4×3×2×1                                                   =5,040   Step 4  Inside the block there are A and B.  The number of ways in which n people can be seated in a row, when there are no restrictions =n!  Therefore, the number of ways in which A and B can be seated inside the block =2!=2  Then, the total number of ways in which 8 people can be seated in a row with A and B must sit together is  The total number of ways =2×5040=10,080ways

Beverly Smith · Accepted Answer

Step 5    (c) To find the number of possible ways in which 8 people can be seated in a row, when there are 4 men and 4 women and no 2 men or 2 women can sit next to each other.    On the first position consider any of the 8 persons, then for one choice of the person for the first position there are 4 possible choices for the second position among the people of opposite sex.   Then for any particular choice of the first two positions there are 3 choices for the third position among the remaining people of the same sex with the person on the first position.    For the fourth position, there are 3 choices among the remaining people of the same sex with the person on the second position.    For the fifth position, there are 2 choices among the remaining people of the same sex with the person on the first position.    For the sixth position, there are 2 choices among the remaining people of the same sex with the person on the second position.    For the seventh position, there is 1 choice left among the remaining people of the same sex with the person on the first position.    For the eigth position, there is 1 choice left among the remaining people of the same sex with the person on the second position.    Multiply all these possiblities to find the number of ways in arranging the people.    The number of ways =8×4×3×3×2×2×1×1=1152ways    Step 6    (d) To find the number of possible ways in which 8 people can be seated in a row, when there are 5 men and they must sit next to one another.    Consider all the men be a single block and then proceed. Now, there are 3 women and one block consisting of 5 men. Observe that there are 4 positions to permute with.   The number of ways in which n people can be seated in a row, when there are no restrictions =n!    The number of ways         =n!                                                   =4!                                              4×3×2×1                                                    =24    Step 7    Inside the block there are 5 men.    The number of ways in which n people can be seated in a row, when there are no restrictions =n!    Therefore, the number of ways in which 5 men can be seated inside the block =5!=120    Then, the total number of ways is    The number of ways =24×120=2880ways

nick1337 · Accepted Answer

Step 8  (e) To find the number of possible ways in which 8 people can be seated  in a row, when there are 4 married couples and each couple must sit  together

Don Sumner · Accepted Answer

a) In the first case, where there are no restrictions on the seating arrangement, the number of ways to seat 8 people in a row can be calculated using the factorial function. The factorial of a number is the product of all positive integers less than or equal to that number. Therefore, the number of ways to arrange 8 people is given by 8! (read as &#039;8 factorial&#039;), which is equal to 40,320.b) In this case, persons A and B must sit next to each other. We can treat persons A and B as a single entity, so we have 7 entities to arrange: {AB, C, D, E, F, G, H}. Now, the number of ways to arrange these 7 entities is 7!. However, within the entity {AB}, persons A and B can be arranged in 2! ways. Therefore, the total number of arrangements is 7! * 2!. Using the factorial notation, we can write it as:7!×2!=10,080.c) In this case, we have 4 men and 4 women, and no two men or two women can sit next to each other. Let&#039;s consider the arrangement of men and women separately.For the 4 men, they must be seated in such a way that no two men sit together. This can be visualized as having 4 spaces between them where women can be seated. We can represent this arrangement as:_M M M M _The 4 men can be seated in these 5 spaces in 5! ways. However, within each group of men, they can be rearranged in 4! ways. Therefore, the total number of arrangements for the men is 5! * 4!.Similarly, for the 4 women, they can be seated in the 4 remaining spaces in 4! ways.To obtain the total number of arrangements, we multiply the number of arrangements for the men by the number of arrangements for the women:5!×4!×4!=1,152.d) In this case, there are 5 men who must sit next to each other. We can treat the group of 5 men as a single entity. Now, we have 4 entities to arrange: {M, M, M, M, M, ABCDEFGH}. The 4 entities can be arranged in 4! ways. However, within the group of 5 men, they can be rearranged in 5! ways. Therefore, the total number of arrangements is 4! * 5!:4!×5!=2,880.e) In this case, there are 4 married couples, and each couple must sit together. We can treat each married couple as a single entity. Now, we have 4 entities to arrange: {AB, CD, EF, GH}. These 4 entities can be arranged in 4! ways. However, within each entity, the couples can be rearranged in 2! ways. Therefore, the total number of arrangements is 4! * 2! * 2! * 2! * 2!:4!×2!×2!×2!×2!=384.

RizerMix · Accepted Answer

a) If there are no restrictions on the seating arrangement, we can simply arrange the 8 people in a row. The number of ways to arrange them is given by the factorial of 8:8!=8×7×6×5×4×3×2×1=40,320b) If persons A and B must sit next to each other, we can treat them as a single entity. Now, we have 7 entities to arrange, where one of them is the pair (A, B). The number of ways to arrange them is:7!×2!=10,080c) If there are 4 men and 4 women, and no two men or two women can sit next to each other, we can think of the arrangement as alternating between men and women. We have two options to start with: either a man or a woman. Let&#039;s consider the case where we start with a man.First, we arrange the men in a row. There are 4 men, so we have 4! ways to arrange them. Next, we arrange the women in the spaces between the men. We have 5 spaces between the 4 men, as well as two ends. We need to choose 4 spaces for the women, so we have 5 choose 4 ways to do this. Finally, we arrange the women within the chosen spaces, which gives us 4! ways.Putting it all together, the total number of arrangements is:4!×(54)×4!=1,152d) If there are 5 men and they must sit next to one another, we can treat the group of 5 men as a single entity. Now, we have 4 entities to arrange, where one of them is the group of 5 men. The number of ways to arrange them is:4!×5!=2,880e) If there are 4 married couples and each couple must sit together, we can treat each couple as a single entity. Now, we have 4 entities to arrange, where each entity represents a couple. The number of ways to arrange them is:4!So the total number of ways to arrange the 8 people in this scenario is 384.

In how many ways can 8 people be seated in a row if a.) there are no r

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