+36 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:
+36916 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
+128474 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 )
+36916 Hey CPhil.... What about the other math book? 5 history books + 1 math book = 6 books in the middle.
+36 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.)
+128474 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 !!! ]
