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# All of the stocks on the OTC market are designated by either

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Manager
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All of the stocks on the OTC market are designated by either [#permalink]

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08 Nov 2005, 22:25
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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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08 Nov 2005, 23:13
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
Senior Manager
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08 Nov 2005, 23:53
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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09 Nov 2005, 00:06
26^4 + 26^5 = 26^4(1+26) = 27(26^4)

C
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09 Nov 2005, 00:08
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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09 Nov 2005, 05:05
This is strange, I am getting:
26^4+26^5 =( 26!/22!*4!) +(26!/21!*5!)=27^5.
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09 Nov 2005, 05:34
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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09 Nov 2005, 08:21
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)

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hey ya......

Manager
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10 Nov 2005, 05:17
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!),
Manager
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10 Nov 2005, 05:22
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)
10 Nov 2005, 05:22
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