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13 Nov 2018, 00:05
Kudos
Bookmarks
Hi,
Could someone help clarify my P&C understanding based on this questions
If there are 3 boys and 3 girls, then in how many ways can they be arranged such that.......
(a) 3 girls cannot be together
(b) 2 girls cannot be together
(c) girl cannot be between boy
(d) girl cannot be between boy, and boy cannot be between girl
(a) Group 3 girls together, so total ways: 4! X 6(no. of ways to arrange 3 girls within the group) = 144
(b) choose 2 girls out of 3 girls = 3 ways. Next, group the 2 girls together, so we now have 5 group in total -> No. of ways = 5!*3(no.of ways to choose 2 out of 3 girls)* 2 (no. of ways to arrange the 2 girls within the group)= 720 ways. Next, calculate no of ways when the 3 girls are together -> no. of ways = 144. Therefore, ans: 6!(total no of ways to arrange without any restriction) - 720 (2 girls together) - 144 (3 girls together) [surely this answer is wrong]
(c) Group 2 boys and 1 girl together= 3 choose 2 X 3 Choose 1 = 3 X 3 = 9 ways to form the group. Within this group, there are 6 ways to arrange them. Therefore, ans: 6!(total no of ways to arrange without any restriction) - (9X6)
d) I am confused with this since i assume (d) arrangement is also considered in the arrangement stated in part (c) and thus (d) and (c) should have the same answer?
Can someone help clarify my understanding for (b), (c), (d). I understand that for (b), i can use the slot method where *B*B*B, but I would like to know how to solve it with the method I described above
Appreciate all help!!!
Could someone help clarify my P&C understanding based on this questions
If there are 3 boys and 3 girls, then in how many ways can they be arranged such that.......
(a) 3 girls cannot be together
(b) 2 girls cannot be together
(c) girl cannot be between boy
(d) girl cannot be between boy, and boy cannot be between girl
(a) Group 3 girls together, so total ways: 4! X 6(no. of ways to arrange 3 girls within the group) = 144
(b) choose 2 girls out of 3 girls = 3 ways. Next, group the 2 girls together, so we now have 5 group in total -> No. of ways = 5!*3(no.of ways to choose 2 out of 3 girls)* 2 (no. of ways to arrange the 2 girls within the group)= 720 ways. Next, calculate no of ways when the 3 girls are together -> no. of ways = 144. Therefore, ans: 6!(total no of ways to arrange without any restriction) - 720 (2 girls together) - 144 (3 girls together) [surely this answer is wrong]
(c) Group 2 boys and 1 girl together= 3 choose 2 X 3 Choose 1 = 3 X 3 = 9 ways to form the group. Within this group, there are 6 ways to arrange them. Therefore, ans: 6!(total no of ways to arrange without any restriction) - (9X6)
d) I am confused with this since i assume (d) arrangement is also considered in the arrangement stated in part (c) and thus (d) and (c) should have the same answer?
Can someone help clarify my understanding for (b), (c), (d). I understand that for (b), i can use the slot method where *B*B*B, but I would like to know how to solve it with the method I described above
Appreciate all help!!!
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
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14 Nov 2018, 07:27
Kudos
Bookmarks
Crusaders
Hi,
I think you are wrong in A alaoy..
So..
A) 3 girls cannot be together..
Total ways = 6!
Ways all three are together take all three as 1, so 3 boys and 1 group..
Choose one place out of 4 for group = 4C1=4..
In each the girls can be arranged in 3! Ways so 4*3!
Answer = 6!-4*3!=6!-4!
B) 2 girls cannot be together
So place boys _B_B_B_ the girls can be seated in any of these 4 seats
so 4C3*3!*3!
C). Girls cannot be between boys
So all 3 boys together..
Same as in A we had found all girls together
4*3!
D) girl cannot be between boys and boys between girls..
So both are separate group 2 groups B and G.. so BG and GB...2 ways
Within themselves 3! Each
So 2*3!3!
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
