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Combination problem

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by jayhawk2001 » Thu Feb 22, 2007 10:14 pm
Max number of 4 letter stocks = 26 * 26 * 26 * 26 = 26^4

Max number of 5 letter stocks = 26^5

Hence max number of 4 or 5 letter stocks = 26^4 + 26^5 = 27*(26^4)
which is option C
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by Stacey Koprince » Mon Feb 26, 2007 11:04 pm
Just wanted to spell out jayhawk's last step a little further:

26^4 + 26^5 = 26^4(1 + 26) = 27*(26^4)
(so, you're pulling out the common term, 26^4, to simplify)
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by f2001290 » Mon May 28, 2007 3:33 am
Jay

Could you please help me with this prob. question?

No of 4 letter words = 26C4*4!
No of 5 letter words = 26c5 * 5!

Total no of words = 26c4*4! + 26C5*5!

What is wrong with this? Probability is a pain..
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by Prasanna » Mon May 28, 2007 5:15 am
f2001290 wrote:Jay

Could you please help me with this prob. question?

No of 4 letter words = 26C4*4!
No of 5 letter words = 26c5 * 5!

Total no of words = 26c4*4! + 26C5*5!

What is wrong with this? Probability is a pain..
I agree with f2001290. The question has confused me. I was trying to solve it through Permutation :shock:
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by f2001290 » Mon May 28, 2007 6:25 am
I got the mistake with my approach.

If a 4 letter code is AAAA. 26C4*4! doesn't take this in to consideration.

Similar is the case with 5 letter words.
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by jayhawk2001 » Mon May 28, 2007 8:10 am
f2001290 wrote:I got the mistake with my approach.

If a 4 letter code is AAAA. 26C4*4! doesn't take this in to consideration.

Similar is the case with 5 letter words.
Yes. Just to clarify,

choices for choosing 1st alphabet = 26
choices for choosing 2nd alphabet = 26
...etc

If you are doing 26C4, you are essentially excluding the alphabet that
you picked first for the 2nd, 3rd and 4th position.
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