Can someone please please solve this problem?Ü?

Question

It's on permutation.. In how many ways can 4 boys and 3 girls be seated in a row of 7 chairs if a) they can sit anywhere? b.)the boys and girls are to be seated alternately?

2.) In how many ways can 9 person be seate3d in a round table with 9 chairs if two individuals insist on sitting next to each other?

*Can you please show me the equations*Ü thanks!

This is on combination...Ü
1.) In a given examination, a student is required to answer 5 of the 7 questions including 3f of the first 4 questions. In how many ways can he anwer the examination?

2.) A multiple choice type of examination is given to a class. Each number has four choices where only one is correct. If there are ten questions, then

a.) in how many ways can a student answer the examination?

b.) in how many ways can the student perfect the examination?

Answer 1

The first person is a boy.
4 possibilities.
The second is a girl: 3 possibilities
So far 4x3 = 12 possibilities.
The third person is one of the left 3 boys:
12 x 3 = 36 possibilities.
Now one of the 2 remaining girls: 36 x 2 =72 possibilities.
Now one of the 2 remaining boys: 72 x 2 = 144 possibilities.
The last girl and boy can be done only at one way.
144 x 1. So there are 144 possibilitie or 144 permutations.

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Answer 2

Problem 1a is the same as asking to seat 7 persons on 7 chairs.
Person 1 can choose among the 7 chairs to sit down on.
Then, independent of where person 1 sits, person 2 can choose among the 6 empty chairs to sit down on (he/she doesn't want to sit on person 1's lap).
This reasoning continues until all persons are seated.
As such, there are 7*6*5*4*3*2*1 = 7! = 5040 possible arrangements.

For problem 1b, label the chairs with "B" if a boy has to sit there, and "G" if a girl has to sit there.
The only possible arrangement of chairs is then: B G B G B G B.
Repeat the reasoning of problem 1a for the 4 boys, and the 3 girls.
The boys can occupy the seats labelled "B" in 4*3*2*1 = 4! = 24 ways, and the girls can occupy the seats labelled "G" in 3*2*1 = 3! = 6 ways.
Since the boys and girls can be seated independent of each other, the 7 persons can be seated in 24*6 = 144 possible ways without violating the imposed restriction.

Problem 2 consists of two independent parts.
First, seat the two persons who wish to sit down next to each other, and then seat the 7 others.
To seat the "couple", it doesn't matter where the first person of this couple sits, since the table is round.
Then, the second person of the couple has 2 choices where to sit, to the left or right of his/her buddy.
To seat the other 7 people, treat the empty seats as standing in a row, and from problem 1a you know that those 7 can be seated in 5040 possible arrangements.
Since the couple and the 7 others are seated independently of each other, the group of 9 can be seated at this round table in 2*5040 = 10080 possible ways without violating the restriction.

Answer 3

1. a) if they can seat any where then its first place can be filled by any of 7. after that second place by left 6 people in 6way .thus total permutation is 7*6*5*4*3*2*1=5040
b) if girl and boys seat alternately then 4 places can be filled up like previous one i.e.4*3*2*1=24 ways and for each of these ways remaining 3 places can be filled by 3*2*1=6 ways.so total permutation is (4*3*2*1)*(3*2*1) =144
2. in round table case first place can be filled by 9 way. as two specific person can't seat side by side so beside two places can be filled by 7 and 6 way (kepping the specified second person away, remaining persons r 7, so one chair by 7 way, as it will fill otherside chair by 6 ways) and the other places can be filled by remaining 6 people in way as in problem 1. so total permutation is 9*7*6*6*5*4*3*2*1=272160

Answer 4

a)first person has 7 choices, the next one 6 and so on.
so the total no. of ways are 7!
b)firstly a boy has to sit.he will have 4 choices. the next boy will have 3 choices and so on.similarly, first girl will have 3 choices, second one will have 2 and the third girl will have 1 choice.so total no. of ways are 4!3!
c)let us consider those two people as 1. now the possible ways are 8!. since, these two persons can also interchange, total no. of ways are 2 x 8!

Answer 5

1a)
4+3=7 person.
7!=1.2.3.4.5.6.7=5040 ways.

b)first boy than girl
4.3.3.2.2.1.1=144 ways


2)
9-2=7
think two individuals as one person
7+1=8
n=8
(8-1)!=7!=1.2.3.4.5.6.7=5040 ways.

Answer 6

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