This math question

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1

<p>We arrange five cards on the table (A, B, C, D, E) such that card C never stands at either end. How many way can we arrange them?</p>

2

<p>Is this a multiple choice?</p>

<p>And is the answer 96? (I’m probably wrong but if not, I can explain)</p>

3

<p>First remove C, and see how many possibilities are there for ABCD. This gives you 4 x 3 x 2 x 1 = 24. Since C can only be put in three spots this means that at every spot there are 24 possibilities for the other cards to be in. So it’ll be 24 + 24 + 24 = 72. Am I right on the answer?</p>

4

<p>There 3 possibilities for C ,4 possibilities for A ,3 possibilities for B ,2 possibilities for D and 1 for C
3x4x3x2 = 72</p>

5

<p>72 is the right one. But I can’t understand why we have 4! ? Where is E? Please elaborate on your answer</p>

6

<p>It’s not 4!. 4! would be 4x3x2, but we’re doing 4x3x3x2.</p>

<p>Start with the two slots with limitations. Normally, there would be 5 options for the first slot and 4 for the last, but since C cannot be in either of these slots there are 4 and 3 options respectively. C is then thrown into the mix, given us 3 options again for the next slot and 2 and 1 options for the last two. 4x3x3x2=72.</p>

7

<p>Or, take the total number of ways of arranging them (5!) and subtract all the ways that have C in the first slot (4!) or last slot (4! again). You’ll get 120 - 24 - 24, which comes out to 72.</p>

8

<p>112358’s method is the perfect standard method. Learn it.</p>

9

<p>Yeah that’s how you do it. </p>

<p>I just imagined it like:
Normally (without the clause that C can’t be on the ends), the number of possibilities is 5! or 5x4x3x2x1.
Now imagine C as the first card that rearranges so there is only 3 possible spots for it to be (5-2 = 3 as it can’t be on the sides). So total possible solutions is 3x4x3x2x1 = 72.</p>

10

<p>how come there are 4! slot when C at the end?</p>

11

<p>when C is fixed at one end… 4 spots are left, and ABCD can be filled in 4<em>3</em>2*1 = 4! ways</p>

12

<p>Check out 476/18 in <a href=“http://talk.collegeconfidential.com/sat-preparation/339734-consolidated-list-blue-book-math-solutions-3rd-ed.html[/url]”>http://talk.collegeconfidential.com/sat-preparation/339734-consolidated-list-blue-book-math-solutions-3rd-ed.html&lt;/a&gt;.&lt;/p&gt;

13

<p>5!-2*4!=72
I guess this is a simpler way^_^</p>