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PostPosted: Fri, 27 Apr 2012 12:12:12 UTC 
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1. There are ten players in a basketball team. The team decides to have a group photo taken and they are supposed to arrange themselves in two rows of five. Find the different number of arrangements if the captain must stand between the two youngest players in the same row.

My answer = 8! X 2 = 80640.

(Group the captain and youngest players together as one unit, 2 ways of arranging youngest players, 8! ways of arranging 8 units.)

But the answer given is 60480. Similar digits, so I was wondering if it's just a printing error? Or am I wrong?

Also, I want to ask: does the fact that they are arranged in two rows affect the number of arrangements? Personally, I don't think so, but I want to check. Thanks.

2. The digits of the number 3215463 are rearranged. If the first and last digits must be different, find how many such numbers must be odd.

My method: find no. of arrangements where last digit is odd (1, 3, 3, 5) and subtract number of arrangements where first and last digits are the same (where number starts with 3 and ends with 3).

Answer = 4C1 X 6! - 5! = 2760.

But the answer given is 1320. Can someone help me out here!

Also, I want to ask - when you have non-distinct entities (in this case 3 and 3) do you still put 4 choose 1 when selecting a last digit that is odd? Because 3 and 3 are essentially identical.

Thanks!


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PostPosted: Fri, 27 Apr 2012 12:19:52 UTC 
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Location: On this day Taiwan becomes another Tiananmen under Dictator Ma.
thesocialnetwork wrote:
1. There are ten players in a basketball team. The team decides to have a group photo taken and they are supposed to arrange themselves in two rows of five. Find the different number of arrangements if the captain must stand between the two youngest players in the same row.

My answer = 8! X 2 = 80640.

(Group the captain and youngest players together as one unit, 2 ways of arranging youngest players, 8! ways of arranging 8 units.)

But the answer given is 60480. Similar digits, so I was wondering if it's just a printing error? Or am I wrong?

Also, I want to ask: does the fact that they are arranged in two rows affect the number of arrangements? Personally, I don't think so, but I want to check. Thanks.

2. The digits of the number 3215463 are rearranged. If the first and last digits must be different, find how many such numbers must be odd.

My method: find no. of arrangements where last digit is odd (1, 3, 3, 5) and subtract number of arrangements where first and last digits are the same (where number starts with 3 and ends with 3).

Answer = 4C1 X 6! - 5! = 2760.

But the answer given is 1320. Can someone help me out here!

Also, I want to ask - when you have non-distinct entities (in this case 3 and 3) do you still put 4 choose 1 when selecting a last digit that is odd? Because 3 and 3 are essentially identical.

Thanks!


1. In using 8!*2, you included cases where the three could sit in different rows but will be "together" when you flatten 2*5 to 1*10.

2. You counted those ending in 3 twice with (4 choose 1)*6!

_________________
\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: Fri, 27 Apr 2012 12:50:34 UTC 
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thanks! (:


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PostPosted: Fri, 27 Apr 2012 16:48:49 UTC 
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Joined: Mon, 19 May 2003 19:55:19 UTC
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Hello, thesocialnetwork!

Quote:
1. There are ten players in a basketball team. The team decides to have a group photo taken
and they are supposed to arrange themselves in two rows of five. Find the different number
of arrangements if the captain must stand between the two youngest players in the same row.

My answer = 8! X 2 = 80640.

(Group the captain and youngest players together as one unit,
. . 2 ways of arranging youngest players, 8! ways of arranging 8 units.)

But the answer given is 60480.

Your approach is quite good . . .

Let the captain be C.
Let the two youngest players be A and B.


We have: .\boxed{AC\!B} \text{ or }\boxed{BC\!A} . . . . two ways.


This group can be seated in six ways:

\begin{array}{ccccccccc} 
\boxed{\circ\;\circ\;\circ}\;\circ\;\circ &&  \circ\;\boxed{\circ\;\circ\;\circ}\;\circ && \circ\;\circ\;\boxed{\circ\;\circ\;\circ} \\ \circ\;\circ\;\circ\;\circ\;\circ &&  \circ\;\circ\;\circ\;\circ\;\circ && \circ\;\circ\;\circ\;\circ\;\circ \end{array} \quad\;\; \begin{array}{ccccccccc} 
\circ\;\circ\;\circ\;\circ\;\circ &&  \circ\;\circ\;\circ\;\circ\;\circ && \circ\;\circ\;\circ\;\circ\;\circ \\ \boxed{\circ\;\circ\;\circ}\;\circ\;\circ &&  \circ\;\boxed{\circ\;\circ\;\circ}\;\circ && \circ\;\circ\;\boxed{\circ\;\circ\;\circ} \end{array}


The remaining players can be seated in 7! ways.


Answer: .2\cdot 6 \cdot 7! \:=\:60,480



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PostPosted: Fri, 27 Apr 2012 17:11:34 UTC 
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yes i realized that, i used your approach when i realized my mistake (: thanks for taking the time to list them all out!


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