SOLUTION: Four girls and three boys are seated randomly in a row. Calculate the following probabilities. (a) The girls occupy the first three places. (b) They are arranged to alternate gir

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Question 1194074: Four girls and three boys are seated randomly in a row. Calculate the following probabilities.
(a) The girls occupy the first three places.
(b) They are arranged to alternate girl, boy, girl, etc.
Answers
(a) 4/35
(b) 1/35
Can someone explain the answers? Thank you.

Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
Four girls and three boys are seated randomly in a row. Calculate the following probabilities.
(a) The girls occupy the first three places.
(b) They are arranged to alternate girl, boy, girl, etc.
Answers
(a) 4/35
(b) 1/35
Can someone explain the answers? Thank you.
~~~~~~~~~~~~~~~ 

(a)  The total number of all possible different permutations/arrangements, without any constraints, is

         (4+3)! = 7! = 1*2*3*4*5*6*7.


     If the girls occupy first three position, then the number of all such permutations/arrangements is

         (4*3*2) * (7-3)! = (4*3*2) * (4*3*2*1).


     The probability under question (a) is the ratio of these numbers

         P = {{({(4*3*2)*(4*3*2*1))/(1*2*3*4*5*6*7)}}} = .    ANSWER




(b)  There is only one placing scheme, satisfying conditions (b). It is  (G B G B G B G).


     Inside this scheme, 4! * 3! arrangements are possible.


     In all, there are 7! arrangements of 4 girls and 3 boys.


     The ratio of these two numbers is the sough probability  P = .


     Reduce this fraction and get the answer  .

Solved.



Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


The solution from the other tutor is one standard way of answering these two questions. You should understand it and be able to use it to solve similar problems.

Another standard way of solving these problems is to consider the probabilities of filling the seats one at a time under the given constraints.

(1) girls in first three places...

first place: 7 people to choose from; 4 are girls --> P = 4/7
second place: 6 people to choose from; 3 are girls --> P = 3/6
third place: 5 people to choose from; 2 are girls --> P = 2/5

P(girls in first three places) = (4/7)(3/6)(2/5) = 4/35

(2) alternating...

Obviously a girl must be first, then a boy, and so on. Without showing in words the probabilities for getting the right gender in the right seat...

P(alternating) = (4/7)(3/6)(3/5)(2/4)(2/3)(1/2)(1/1) = 1/35
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