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More questions.. Halp!

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1)  In how many ways can I arrange 3 different math books and 5 different history books on my bookshelf, if I require there to be a math book on both ends?

2)  I have 10 distinguishable socks in my drawer: 4 white, 4 brown, and 2 blue. In how many ways can I choose a pair of socks, provided that I get two socks of different colors?

tytyty! c:

Feb 17, 2018

#1
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Dont post problems from AOPS, its not worth it

Feb 17, 2018
#3
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Your keyboard seems to have a problem with the apostrophe key.

ElectricPavlov  Feb 18, 2018
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I think:

1.

There are 6 different ways a math book can be at both ends.

That leaves 6 books in the middle that can be arranged 6! ways

6 x 6! = 4320

Feb 18, 2018
#4
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1)  In how many ways can I arrange 3 different math books and 5 different history books on my bookshelf, if I require there to be a math book on both ends?

We have   P(3,2)  =  6 ways to arrange the math books  and 6! ways of arranging the 1 math book and 5 history books in the middle....  so...6 x 6!  =   6 x 720  =  4320 ways

2)  I have 10 distinguishable socks in my drawer: 4 white, 4 brown, and 2 blue. In how many ways can I choose a pair of socks, provided that I get two socks of different colors?

We  can  either have

White - Brown  =  4C1 x 4C1  =  4 x 4  =  16 ways

Brown - Blue  =  4C1 x 2C1   =   4 x 2   = 8  ways

White -Blue  = 4C1  x 2C1  =   4 x 2  =  8 ways

So...the total ways =   16 + 8 + 8  =   32

( EDIT )   Feb 18, 2018
edited by CPhill  Feb 18, 2018
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Hey CPhil....   What about the other math book?  5 history books + 1 math book = 6 books in the middle.

ElectricPavlov  Feb 18, 2018
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You are correct, EP!!!!...I'll edit my answer  !!!   CPhill  Feb 18, 2018
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Hi, I don't understand how you got the p(3,2) can you explain it to me? (Thank you! I'm really struggling with this topic.)

Echotastic  Feb 18, 2018
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P (3,2)   means that  we are choosing any 2 things out of 3 and then permuting (giving all the possible orders  of the things selected )

So...suppose the math books  are  A, B and C

The possible selections we can make is

{ A, B}   { B , C}  or { A, C}    (1)

But  we can also arrange these as  { B, A}  {C, B}  or   {C, A}.......so we have a total of 6 possible arrangements

Note that   (1)  is  C(3,2) = 3  =   the number of possible sets

But  P(3,2)   =  6  =  the number of posssible arrangements of these sets

[ This stuff is hard for everyone  !!!  ]   Feb 18, 2018