Thread: permutation and combination
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Old June 25th, 2008, 09:47 AM
Soroban Soroban is offline
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Hello, divinewisdom0218!

Quote:
1. Six books with different titles have green covers
and three books with different titles have red covers.
In how many ways can the nine books be arranged on a shelf if:
(a) the red books are not to be separated,
(b) the green books are not to be separated,
(c) books of the same color are to be kept together?

(a) Red books are together.

Tape the 3 red books together.
Then we have 7 "books" to arrange: .\boxed{RRR},\:G,\:G,\:G,\:G,\:G,\:G

These can be arranged in 7! ways.
. . But the 3 red books can be ordered in 3! ways.

Therefore, there are: .(7!)(3!) \:=\:30,240 arrangements.


(b) Green books are together.

Tape the 6 green books together.
Then we have 4 "books" to arrange: .R,\:R,\:R,\:\boxed{GGGGGG}

These can be arranged in 4! ways.
. . But the 6 green books can be arranged in 6! ways.

Therefore, there are: .(4!)(6!) \:=\:17,280 arrangements.


(c) Greens together, reds together.

Tape the green books together; tape the red books together.
Then we have 2 "books" to arrange: .\boxed{GGGGGG},\:\boxed{RRR}

These can be arranged in 2! ways.
But the 6 green books can be ordered in 6! ways.
. . and the 3 red books can be ordered in 3! ways.

Therefore, there are: .(2!)(6!)(3!) \:=\:8,640 arrangements.



Quote:
2. If a man has 3 suits, 4 shirts, 10 ties, 6 pairs of socks, 2 hats,
and 3 pairs of shoes, in how many ways can he be dressed?

Assuming that he chooses one of each item of clothing,
. . there are: .3 \times 4 \times 10 \times 6 \times 2 \times 3 \:=\:4,320 ways.
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