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PostPosted: Sat, 28 Apr 2012 09:47:25 UTC 
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1. Find the number of different ways in which 5 women and 6 men can stand in a line if 4 or more women stand next to one another.

My approach: add no. of ways where exactly 4 women stand next to one another + no. of ways where exactly 5 women stand next to one another

= 7C2 x 2! x 4! x 6! + 7C1 x 5! x 6!
= 1330560

But the answer given is 4233600. Can someone tell me where I went wrong?

2. Find how many 4-digit numbers can be formed using only the digits 3,4,6 and 8, no digit being repeated and if the numbers are
i) divisible by 2
ii) divisible by 4

answer to i) is 18 and ii) is 10. I have i) but I don't know how to get ii) - I don't think it's a P and C problem haha, can someone tell me how you get a number that is divisible by 4 in this case? Does the number have to end with a four etc? Thanks.

i) is 3C1 x 3! = 18, but I don't know how to get ii).

Thank you!


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PostPosted: Sat, 28 Apr 2012 10:04:13 UTC 
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thesocialnetwork wrote:
1. Find the number of different ways in which 5 women and 6 men can stand in a line if 4 or more women stand next to one another.

My approach: add no. of ways where exactly 4 women stand next to one another + no. of ways where exactly 5 women stand next to one another

= 7C2 x 2! x 4! x 6! + 7C1 x 5! x 6!
= 1330560

But the answer given is 4233600. Can someone tell me where I went wrong?

2. Find how many 4-digit numbers can be formed using only the digits 3,4,6 and 8, no digit being repeated and if the numbers are
i) divisible by 2
ii) divisible by 4

answer to i) is 18 and ii) is 10. I have i) but I don't know how to get ii) - I don't think it's a P and C problem haha, can someone tell me how you get a number that is divisible by 4 in this case? Does the number have to end with a four etc? Thanks.

i) is 3C1 x 3! = 18, but I don't know how to get ii).

Thank you!


1. Where does the 7C2 x 2! x 4! x 6! come from? To get exactly 4 women next to each other, you have either they are in the middle, in which case you want two men either side, or they are at one end of the line, in which case next to these 4 women must be a man, so ... Or think about inclusion-exclusion. And as a further reason why your answer must be wrong, note that the answer must be divisible by 5!6! (permuting the men, and permuting the women, separately, must give different permutation of the line).

2. Such a number is divisible by 4 if and only if it ends in 36, 48, 64, 68 or 84 (recall divisibility by 4 is determined by the last 2 decimal digits).

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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PostPosted: Sat, 28 Apr 2012 10:17:12 UTC 
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For 1. I slotted [W W W W] and [W] into one of the [ ] below, so that's 7 choose 2. Then two ways of arranging [W W W W] and [W] and 4! ways of arranging the women in [W W W W] and finally 6! ways of arranging the 6 men. This achieves the scenario of exactly four women standing next to one another and the next part achieves the scenario where exactly 5 women stand next to each other.

[ ] M [ ] M [ ] M [ ] M [ ] M [ ] M [ ]


by inclusion-exclusion do you mean calculate the ways in which 0, 1, 2 or 3 women stand next to each other and subtract them from the total? that's possible but i thought it would be rather more tedious (say under exam conditions) and i don't understand how my above method is wrong.

thanks for your help!


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PostPosted: Sat, 28 Apr 2012 13:57:28 UTC 
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thesocialnetwork wrote:
For 1. I slotted [W W W W] and [W] into one of the [ ] below, so that's 7 choose 2. Then two ways of arranging [W W W W] and [W] and 4! ways of arranging the women in [W W W W] and finally 6! ways of arranging the 6 men. This achieves the scenario of exactly four women standing next to one another and the next part achieves the scenario where exactly 5 women stand next to each other.

[ ] M [ ] M [ ] M [ ] M [ ] M [ ] M [ ]


by inclusion-exclusion do you mean calculate the ways in which 0, 1, 2 or 3 women stand next to each other and subtract them from the total? that's possible but i thought it would be rather more tedious (say under exam conditions) and i don't understand how my above method is wrong.

thanks for your help!


You forgot the 5! permutation of the women's positions.

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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PostPosted: Sat, 28 Apr 2012 15:49:31 UTC 
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Hm but i've already taken the arrangement of the women into account when i permute: 2 ways of arranging groups [W W W W] and [W] and 4! ways of arranging the 4 women within the group [W W W W].


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PostPosted: Sat, 28 Apr 2012 15:58:55 UTC 
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thesocialnetwork wrote:
Hm but i've already taken the arrangement of the women into account when i permute: 2 ways of arranging groups [W W W W] and [W] and 4! ways of arranging the 4 women within the group [W W W W].


No, you haven't taken into account ways of selecting which woman to be left out from the group of 4.

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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PostPosted: Sat, 28 Apr 2012 17:42:06 UTC 
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ah yes thank you!!


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PostPosted: Sun, 4 Aug 2013 08:15:26 UTC 
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outermeasure wrote:
thesocialnetwork wrote:
For 1. I slotted [W W W W] and [W] into one of the [ ] below, so that's 7 choose 2. Then two ways of arranging [W W W W] and [W] and 4! ways of arranging the women in [W W W W] and finally 6! ways of arranging the 6 men. This achieves the scenario of exactly four women standing next to one another and the next part achieves the scenario where exactly 5 women stand next to each other.

[ ] M [ ] M [ ] M [ ] M [ ] M [ ] M [ ]


by inclusion-exclusion do you mean calculate the ways in which 0, 1, 2 or 3 women stand next to each other and subtract them from the total? that's possible but i thought it would be rather more tedious (say under exam conditions) and i don't understand how my above method is wrong.

thanks for your help!


You forgot the 5! permutation of the women's positions.


This original thread is old, but I came across it and was doing this question with my engineer nephew. He thinks it is unnecessary to worry about the possibilities for WWWW and W, since they are contained within the 5!. I cannot make up my mind whether he is correct, but following his idea we get the answer that the original poster said was the supposed answer. Is he right or not right?


Last edited by salsero on Sun, 4 Aug 2013 08:22:35 UTC, edited 1 time in total.

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PostPosted: Sun, 4 Aug 2013 08:17:17 UTC 
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salsero wrote:
outermeasure wrote:
thesocialnetwork wrote:
For 1. I slotted [W W W W] and [W] into one of the [ ] below, so that's 7 choose 2. Then two ways of arranging [W W W W] and [W] and 4! ways of arranging the women in [W W W W] and finally 6! ways of arranging the 6 men. This achieves the scenario of exactly four women standing next to one another and the next part achieves the scenario where exactly 5 women stand next to each other.

[ ] M [ ] M [ ] M [ ] M [ ] M [ ] M [ ]


by inclusion-exclusion do you mean calculate the ways in which 0, 1, 2 or 3 women stand next to each other and subtract them from the total? that's possible but i thought it would be rather more tedious (say under exam conditions) and i don't understand how my above method is wrong.

thanks for your help!


You forgot the 5! permutation of the women's positions.


This original thread is old, but I came across it and was doing this question with my engineer nephew. He things it is unnecessary to worry about the possibilities for WWWW and W, since they are contained within the 5!. I cannot make up my mind whether he is correct, but following his idea we get the answer that the original poster said was the supposed answer. Is he right or not right?


The solution presented here is the correct one.

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PostPosted: Sun, 4 Aug 2013 08:24:56 UTC 
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Shadow wrote:
salsero wrote:
outermeasure wrote:
thesocialnetwork wrote:
For 1. I slotted [W W W W] and [W] into one of the [ ] below, so that's 7 choose 2. Then two ways of arranging [W W W W] and [W] and 4! ways of arranging the women in [W W W W] and finally 6! ways of arranging the 6 men. This achieves the scenario of exactly four women standing next to one another and the next part achieves the scenario where exactly 5 women stand next to each other.

[ ] M [ ] M [ ] M [ ] M [ ] M [ ] M [ ]


by inclusion-exclusion do you mean calculate the ways in which 0, 1, 2 or 3 women stand next to each other and subtract them from the total? that's possible but i thought it would be rather more tedious (say under exam conditions) and i don't understand how my above method is wrong.

thanks for your help!


You forgot the 5! permutation of the women's positions.


This original thread is old, but I came across it and was doing this question with my engineer nephew. He things it is unnecessary to worry about the possibilities for WWWW and W, since they are contained within the 5!. I cannot make up my mind whether he is correct, but following his idea we get the answer that the original poster said was the supposed answer. Is he right or not right?


The solution presented here is the correct one.


Is the solution actually written out clearly somewhere here? I can't see it written out neatly anywhere. And sorry if what is written is not a good enough "solution" for me, but this type of question, or math in general is not my forté, to say the least. We must be doing something wrong if we are getting the right answer when doing something wrong...


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PostPosted: Sun, 4 Aug 2013 08:29:27 UTC 
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We almost never have final neatly written solutions, we're a help board and we talk through things and give hints and such. If you want to see how the solution goes you'll have to follow the map by reading the topic, sorry.

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PostPosted: Sun, 4 Aug 2013 08:39:01 UTC 
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Shadow wrote:
We almost never have final neatly written solutions, we're a help board and we talk through things and give hints and such. If you want to see how the solution goes you'll have to follow the map by reading the topic, sorry.


Okay. Good enough, Shadow. Thank you for your time. I will go back to work on the question. Thank you for clarifying that my nephew's idea is incorrect.


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PostPosted: Sun, 4 Aug 2013 08:47:12 UTC 
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No problem, glad to keep you for waiting unnecessarily. If you like, you can make a new topic with the question you care about and what steps you've tried already as well as where you're unsure and we can give more specific comments. Also, you're welcome to post other questions if you end up with them, similarly including the work you've tried and where you think you're stuck if you happen to have others.

Best,
Shadow

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PostPosted: Sun, 4 Aug 2013 10:04:50 UTC 
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Shadow wrote:
No problem, glad to keep you for waiting unnecessarily. If you like, you can make a new topic with the question you care about and what steps you've tried already as well as where you're unsure and we can give more specific comments. Also, you're welcome to post other questions if you end up with them, similarly including the work you've tried and where you think you're stuck if you happen to have others.

Best,
Shadow


Okay, here is what I did: (5! * 7) + (4! *5) (7*6) = 840 + 5040 = 5880. Finally, multiply by 6! for men, to get 4,233,600.

5! is for the WWWWW arrangements, and the 7 is for the seven possible positions of them.
My 4! is for the WWWW arrangements. My 5 is for the W possibilities. My 7*6 represents placements for WWWW and W.
Is there anything wrong or false about what I did?


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