permutation/combination help please

mstsoy

New member
Joined
May 7, 2006
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2
"ten pennies are lying on a table. four are 'heads' and six are 'tails.' Show that there are 210 ways such as "HHTTHTHTHT" that the coins can be lined up, considering only 'heads' and 'tails.'"
-I am so lost. I can't figure out whether to use permutaion or combination..I thought maybe combination since the order doesn't seem to matter..but I'm not getting 210 as the answer. Any ideas?



a hat contains tickets w/numbers 1,2...20 printed on them. if 4 tickets are drawn from the hat, without replacement, determine the probability that all are primes.

-well there are 8 primes. i did (8x7x6x5)/20P4 and got 14/969
i was wondering if that's correct

Any help will be appreciated thank you
 

daon

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Jan 27, 2006
Messages
1,284
mstsoy said:
"ten pennies are lying on a table. four are 'heads' and six are 'tails.' Show that there are 210 ways such as "HHTTHTHTHT" that the coins can be lined up, considering only 'heads' and 'tails.'"
-I am so lost. I can't figure out whether to use permutaion or combination..I thought maybe combination since the order doesn't seem to matter..but I'm not getting 210 as the answer. Any ideas?



a hat contains tickets w/numbers 1,2...20 printed on them. if 4 tickets are drawn from the hat, without replacement, determine the probability that all are primes.

-well there are 8 primes. i did (8x7x6x5)/20P4 and got 14/969
i was wondering if that's correct

Any help will be appreciated thank you
1) Permutation with repeats: \(\displaystyle \frac{10!}{4!6!} = 210\)
2) Almost.. Remember that when you take one ticket out, there is one less in the hat. \(\displaystyle P = \frac{8}{20} \cdot \frac{7}{19} \cdot \frac{6}{18} \cdot \frac{5}{17} = \frac{1680}{5814}\)
 

TchrWill

Full Member
Joined
Jul 7, 2005
Messages
856
mstsoy said:
"ten pennies are lying on a table. four are 'heads' and six are 'tails.' Show that there are 210 ways such as "HHTTHTHTHT" that the coins can be lined up, considering only 'heads' and 'tails.'"
-I am so lost. I can't figure out whether to use permutaion or combination..I thought maybe combination since the order doesn't seem to matter..but I'm not getting 210 as the answer. Any ideas?

Any help will be appreciated thank you.
The number of permutations of "n" things, taken all at a time, "p" things of one kind, and "q" things of another kind, is nP(pq) = 10!/[(4!)(6!)] 10!/4!6! = 210.
 

mstsoy

New member
Joined
May 7, 2006
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2
thank you very much, i actually understand this now :)
 

soroban

Elite Member
Joined
Jan 28, 2005
Messages
5,588
Hello, mstsoy!

Here's a baby-talk approach to the first problem . . .

Ten pennies are lying on a table. Four are 'heads' and six are 'tails.'
Show that there are 210 ways that the coins can be lined up.
There are ten positions in a row for the coins.

We will chose four of them and place the Heads there.
\(\displaystyle \;\;\)(The six Tails will go in the remaining positions.)

How many ways can we choose 4 positions out of 10?

There are: \(\displaystyle \,C(10,4)\:=\:\frac{10!}{4!\cdot6!}\:=\: 210\) ways . . . ta-DAA!
 
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