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Manager
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All of the stocks on the OTC market are designated by either [#permalink]
 08 Nov 2005, 22:25
All of the stocks on the OTC market are designated by either a 4 - letter or 5- letter code that is created by using the 26 letters of the alphabet. Which of the following gives the maximum number of diff't stocks that can be designated w/ these?
a. 2(26^5)
b. 26(26^4)
c. 27(26^4)
d. 26(26^5)
e. 27(26^5)
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VP
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c.
number of 4 letter codes with repitition:
26^4
number of 5 letter codes with repitition:
26^5
total: 26^4 + 26^5 or 26^4 (1+26) = 27 * 26^4
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Senior Manager
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I think we need to assume that we can have repetitive letters in the codes, i.e. AAAA or AAAAA
26^4 + 26^5 = 26^4 ( 1+ 26) = 26^4 * 27
I pick D.
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26^4 + 26^5 = 26^4(1+26) = 27(26^4)
C
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Director
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rigger wrote: I think we need to assume that we can have repetitive letters in the codes, i.e. AAAA or AAAAA
26^4 + 26^5 = 26^4 ( 1+ 26) = 26^4 * 27
I pick D.
Aren't you really picking C? 
Better not do that on the actual exam! 
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Manager
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This is strange, I am getting:
26^4+26^5 =( 26!/22!*4!) +(26!/21!*5!)=27^5.
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Manager
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rianah100 wrote: This is strange, I am getting:
26^4+26^5 =( 26!/22!*4!) +(26!/21!*5!)=27^5.
I dont get your calculations.. You cannot use factorials here.
26^4 is not equal to the term you wrote in parenthesis.
You wrote 26*26*26*26 = 26*25*24*23/4*3*2*1 ....
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it is not a combination problem, it is a binomial problem.
The reason is that, a single letter can be repeated more than once.
so maximum possible number of stock is 26^4+26^5 = 26^4( 1+26)
= 27(26^4)
So the answer is C.
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hey ya......
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Manager
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nero44 wrote: rianah100 wrote: This is strange, I am getting:
26^4+26^5 =( 26!/22!*4!) +(26!/21!*5!)=27^5. I dont get your calculations.. You cannot use factorials here. 26^4 is not equal to the term you wrote in parenthesis. You wrote 26*26*26*26 = 26*25*24*23/4*3*2*1 .... The value of the binomial coefficient is given explicitly by
_nC_k=(n; k)=(n!)/((n-k)!k!),
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rianah100 wrote: nero44 wrote: rianah100 wrote: This is strange, I am getting:
26^4+26^5 =( 26!/22!*4!) +(26!/21!*5!)=27^5. I dont get your calculations.. You cannot use factorials here. 26^4 is not equal to the term you wrote in parenthesis. You wrote 26*26*26*26 = 26*25*24*23/4*3*2*1 .... The value of the binomial coefficient is given explicitly by
_nC_k=(n; k)=(n!)/((n-k)!k!),[/C(n,k) = C(n-1,k) + C(n-1,k-1)
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