Perms and Combs :( (1 Viewer)

[Astraea]

In how many ways can 4 boys, including John and Matt, and 4 girls, including Sally, be arranged in a line if:

(i) Girls and Boys Alternate?
(ii) Girls and boys alternate and John and Sally stay together?
(iii) John and Sally must be separated by at least 2 people?
(iv) Matthew is between, but not necessarily adjacent to John and Sally?

 

[cineti970128]

I'll just post the answers
i) 2×4!× 4!
ii) 4!× 4!
iii) 8!-2! 7!-2! 6! 6= 21600
iv) 6!× 2! + 5P1×5!× 2! + 5P2×2! 4! 3 + 5P3×2! 3! 4 + 5P4 ×2! 2! 5 + 5P5×2!× 6×1! = 12240

 

[hit patel]

1) There are 4 girls and 4 boys which can be alternated in 2 groups of 4! x 4!. However at each place where a girl can sit a boy can also sit because the number of both are equal so answer is 4! x 4! x 2

2) So here Sally and John can be considered a group and the rest 3 girls and 3 boys must alternate. 3! x 3! x2! x 7 x 2!.
3! x 3! x 2! because boys and girls alternate. Also 2 factor because sally and john in a group can alternate and can be in a group at 7 places in the line. PLEASE CORRECT ME IF I AM WRONG. I think I am wrong at either the alternation 2! part or the 7 part.

3) Not sure if there is combinations involved but there would be I think so (2! x 6C2 x 2! x 4!) +(2! x 6C3 x 3! x 3!) +(2! x 6C4 x 4! x 2!) + (2! x 6C5 x 5!) + (6! x 2!)
Explanation: Since when each person is added inbetween we have to choose them from the remaining 6 there 6Cn. However john and sally can also swap so 2!. Also the people in between can also swap after they have been chosen and the people not inbetween can also swap. Therefore at each stage a person chosen increases therefore more permutation exist inbetween john and sally then outside. The 2! exists in each part because john and sally can always swap. The remaining outsiders can also swap however this decreases in each part.

4) Basically if my last one was right then:
(2! x 5C1x 2! x 4!) +(2! x 5C2 x 3! x 3!) +(2! x 5C3x 4! x 2!) + (2! x 5C4x 5!) + (6! x 2!)

As visible the only difference from last pattern is that in the last question the combination was 6Cn however now its 5C(n-1). This is because one person is already chosen to be in the middle and the total number of people that can be chosen has also decreased.

Note: These questions could have been faster by subtracting total permutations of john and sally being together by permutations that would occur when 0 ,1 people are inbetween.however this was not chosen to make it easier to understand. I recommend you try this method to also get clear understanding of complementary arrangents.

 

[hit patel]

1) There are 4 girls and 4 boys which can be alternated in 2 groups of 4! x 4!. However at each place where a girl can sit a boy can also sit because the number of both are equal so answer is 4! x 4! x 2

2) So here Sally and John can be considered a group and the rest 3 girls and 3 boys must alternate. 3! x 3! x2! x 7 x 2!.
3! x 3! x 2! because boys and girls alternate. Also 2 factor because sally and john in a group can alternate and can be in a group at 7 places in the line. PLEASE CORRECT ME IF I AM WRONG. I think I am wrong at either the alternation 2! part or the 7 part.

3) Not sure if there is combinations involved but there would be I think so (2! x 6C2 x 2! x 4!) +(2! x 6C3 x 3! x 3!) +(2! x 6C4 x 4! x 2!) + (2! x 6C5 x 5!) + (6! x 2!)
Explanation: Since when each person is added inbetween we have to choose them from the remaining 6 there 6Cn. However john and sally can also swap so 2!. Also the people in between can also swap after they have been chosen and the people not inbetween can also swap. Therefore at each stage a person chosen increases therefore more permutation exist inbetween john and sally then outside. The 2! exists in each part because john and sally can always swap. The remaining outsiders can also swap however this decreases in each part.

4) Basically if my last one was right then:
(2! x 5C1x 2! x 4!) +(2! x 5C2 x 3! x 3!) +(2! x 5C3x 4! x 2!) + (2! x 5C4x 5!) + (6! x 2!)

As visible the only difference from last pattern is that in the last question the combination was 6Cn however now its 5C(n-1). This is because one person is already chosen to be in the middle and the total number of people that can be chosen has also decreased.

Note: These questions could have been faster by subtracting total permutations of john and sally being together by permutations that would occur when 0 ,1 people are inbetween.however this was not chosen to make it easier to understand. I recommend you try this method to also get clear understanding of complementary arrangents.

So am I wrong cineti?

 

[Astraea]

I'll just post the answers
i) 2×4!× 4!
ii) 4!× 4!
iii) 8!-2! 7!-2! 6! 6= 21600
iv) 6!× 2! + 5P1×5!× 2! + 5P2×2! 4! 3 + 5P3×2! 3! 4 + 5P4 ×2! 2! 5 + 5P5×2!× 6×1! = 12240

Can you explain how you did the 2nd and 4th one?

 

[braintic]

I agree with Hit Patel's answer for no 2 (ie. cineti's answer is incorrect)
However I disagree with both answers for no 4. Wherever John, Sally and Matthew are placed (say for example in 1st, 5th and 7th places), in one third of those cases Matthew will be between John and Sally. So the answer should be one third of 8! which is 13440.

The logic for no 2 is that there are 7 places to place the couple - 1st/2nd or 2nd/3rd or ... or 7th/8th.
Once they are placed, they can be ordered JS or SJ.
Once this has been done, the positions of the males and females in the line has been determined. There are 3! ways of placing the 3 remaining males in these positions, and the same for the remaining females.
So 7 times 2 times 3! times 3!

 

[Astraea]

I agree with Hit Patel's answer for no 2 (ie. cineti's answer is incorrect)
However I disagree with both answers for no 4. Wherever John, Sally and Matthew are placed (say for example in 1st, 5th and 7th places), in one third of those cases Matthew will be between John and Sally. So the answer should be one third of 8! which is 13440.

The logic for no 2 is that there are 7 places to place the couple - 1st/2nd or 2nd/3rd or ... or 7th/8th.
Once they are placed, they can be ordered JS or SJ.
Once this has been done, the positions of the males and females in the line has been determined. There are 3! ways of placing the 3 remaining males in these positions, and the same for the remaining females.
So 7 times 2 times 3! times 3!

Still don't quite get your explanation of (iv) sorry.

 

[HSC2014]

So am I wrong cineti?

Your third answer is wrong; cineti got it right (btw your method is very tedious and inefficient)

And i got 6720 for my fourth answer; there seems to be varying answers so I won't post my solution

Edit: Q1 & Q2 are fine (forgot to mention)
Edit 2: Nvm Q2 is wrong - but braintic's solution to Q2 is right which i assumed was yours
Edit 3: My ans for Q4 is 6720 x 2 = 13440 (forgot factor of 2) - so braintic on the money once again

 

[braintic]

Still don't quite get your explanation of (iv) sorry.

Ignoring the other 5 people, and just worrying about the ordering of John, Sally and Matthew (wherever they might be in the line):

Matthew could be placed first before Sally and John OR he could be placed between Sally and John OR he could be placed third after Sally and John.
As there is nothing special about any one person, these three cases must occur with equal frequency.
So in 1/3 of all possible arrangements, Matthew will be (somewhere) between the other two.
Since there are 8! total arrangements, then the number of arrangements must be 8!/3 = 13440.

 

[HSC2014]

^ Very clean My way was a lot slower.

 

[Astraea]

Is this from Dr. Du ?

Yes, do you go as well?

 

[cineti970128]

Yeab I did 2nd one wrong I did×8 not7 silly...
The fourth I did my one by
(S,_, J) , 5people - 2! For S and J can swap, 6 groups hence6! 1 for B
(S,_,_, J), 4people - 2! For S and J can swap, 5groups hence 5!, 2 places for B and put one of 5 into one 5p1
(S,_,_,_J), 3people - etc
(S,_,_,_,_J) 2 people
(S,_,_,_,_,_J) 1 people
(S,_,_,_,_,_,_, J)

Hence 2!(6! + 2×5! 5P1 + 3× 4! 5P2 + 4×3! 5P3 + 5×2! 5P4 +6×1! 5P5)= 13440 with my previous answers I made a typing error if you can see

 

[cineti970128]

So am I wrong cineti?

You're not wrong!!
Just my typo
ii) - silly mistake