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All of the bonds on a certain exchange are designated by a 3

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Re: All of the bonds on a certain exchange are designated by a 3 [#permalink]

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New post 13 Apr 2013, 08:39
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rakeshd347 wrote:
All of the bonds on a certain exchange are designated by a 3-letter, a 4-letter, or a 5-letter code that is created by using the 26 letters of the alphabet. Which of the following gives the maximum number of different bonds that can be designate with these codes?

(A) 26(26^3 + 26^4)
(B) 26(26^3 + 26^5)
(C) 27(26^3 + 26^5)
(D) 27(26^3) + 26^5
(E) 26^3 + 27(26^5)


3-letter codes = 26^3;
4-letter codes = 26^4;
5-letter codes = 26^5;

Total = 26^3 + 26^4 + 26^5 = 26^3(1+26) + 26^5 = 26^3*27 + 26^5.

Answer: D.

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Re: All of the bonds on a certain exchange are designated by a 3 [#permalink]

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New post 12 Feb 2014, 08:03
shouldn't the question mention if the alphabets can be repeated or not? OR do we need to deduce it from the question stem?
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Re: All of the bonds on a certain exchange are designated by a 3 [#permalink]

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New post 12 Feb 2014, 08:49
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Re: All of the bonds on a certain exchange are designated by a 3 [#permalink]

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New post 14 Nov 2016, 18:33
Bunuel wrote:
rakeshd347 wrote:
All of the bonds on a certain exchange are designated by a 3-letter, a 4-letter, or a 5-letter code that is created by using the 26 letters of the alphabet. Which of the following gives the maximum number of different bonds that can be designate with these codes?

(A) 26(26^3 + 26^4)
(B) 26(26^3 + 26^5)
(C) 27(26^3 + 26^5)
(D) 27(26^3) + 26^5
(E) 26^3 + 27(26^5)


3-letter codes = 26^3;
4-letter codes = 26^4;
5-letter codes = 26^5;

Total = 26^3 + 26^4 + 26^5 = 26^3(1+26) + 26^5 = 26^3*27 + 26^5.

Answer: D.

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if-a-code-word-is-defined-to-be-a-sequence-of-different-126652.html
the-simplastic-language-has-only-2-unique-values-and-105845.html
a-4-letter-code-word-consists-of-letters-a-b-and-c-if-the-59065.html
a-certain-stock-exchange-designates-each-stock-with-a-86656.html
a-5-digit-code-consists-of-one-number-digit-chosen-from-132263.html
a-company-that-ships-boxes-to-a-total-of-12-distribution-95946.html

P.S. Please check the question and the answer choices when posting. Also, please name and tag the questions properly.



Hi Bunuel,

Please could you explain the formula that you have used here.

Thanks.
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Re: All of the bonds on a certain exchange are designated by a 3 [#permalink]

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New post 27 Nov 2016, 05:45
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Bunuel wrote:
rakeshd347 wrote:
All of the bonds on a certain exchange are designated by a 3-letter, a 4-letter, or a 5-letter code that is created by using the 26 letters of the alphabet. Which of the following gives the maximum number of different bonds that can be designate with these codes?

(A) 26(26^3 + 26^4)
(B) 26(26^3 + 26^5)
(C) 27(26^3 + 26^5)
(D) 27(26^3) + 26^5
(E) 26^3 + 27(26^5)


3-letter codes = 26^3;
4-letter codes = 26^4;
5-letter codes = 26^5;

Total = 26^3 + 26^4 + 26^5 = 26^3(1+26) + 26^5 = 26^3*27 + 26^5.

Answer: D.

Similar questions to practice:
all-of-the-stocks-on-the-over-the-counter-market-are-126630.html
if-a-code-word-is-defined-to-be-a-sequence-of-different-126652.html
the-simplastic-language-has-only-2-unique-values-and-105845.html
a-4-letter-code-word-consists-of-letters-a-b-and-c-if-the-59065.html
a-certain-stock-exchange-designates-each-stock-with-a-86656.html
a-5-digit-code-consists-of-one-number-digit-chosen-from-132263.html
a-company-that-ships-boxes-to-a-total-of-12-distribution-95946.html

P.S. Please check the question and the answer choices when posting. Also, please name and tag the questions properly.



Can somebody please explain the formula used here ?
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Re: All of the bonds on a certain exchange are designated by a 3 [#permalink]

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New post 15 Apr 2018, 10:25
Bunuel wrote:
rakeshd347 wrote:
All of the bonds on a certain exchange are designated by a 3-letter, a 4-letter, or a 5-letter code that is created by using the 26 letters of the alphabet. Which of the following gives the maximum number of different bonds that can be designate with these codes?

(A) 26(26^3 + 26^4)
(B) 26(26^3 + 26^5)
(C) 27(26^3 + 26^5)
(D) 27(26^3) + 26^5
(E) 26^3 + 27(26^5)

FCP - Answers above use the
Fundamental Counting Principle to find
total possible code combinations for each kind/category
of code (3-, 4-, and 5-letter codes)

FCP: If a task can be broken into stages &
Stage 1 can be done in X ways;
Stage 2 can be done in Y ways; and
Stage 3 can be done in Z ways,

Then the total number of ways to do the task
= X * Y * Z

• The "tasks" are to create 3-, 4-, and 5-letter codes.
Each is a separate task.
Each can be broken into stages.

3-letter code's possibilities?

Stages are like "slots" into which letters are placed

• Stages/slots in 3-letter code: ___ ___ ___

First slot: there are 26 letters from which to
choose, hence 26 possibilities


Second slot: also 26 possibilities ("FCP without restriction")*
If using the same letter is not allowed, the prompt will say so

Third slot: also 26 possibilities

• 3-letter codes possible: 26*26*26= \(26^3\)

Possibilities for each type of code

3-letter code has \(26^3\) possibilities

4-letter code's possibilities?
_26_*_26_*_26_*_26_ = \(26^4\)

5-letter code's possibilities? \(26^5\)

Now ADD each set's # of possibilities

Sets are 3-, 4-, and 5-letter codes.
One bond gets only one code.

But the total pool of codes for each bond
consists of possibilities from the 3-, 4-, and 5-letter sets

ADD sets' possibilities: \(26^3 + 26^4 + 26^5\) = ?

Rearrange answer to match answer choices
\(26^3 + 26^4 + 26^5\)

Often, we would factor out \(26^3\):
\(26^3(26^1 + 26^2)\)


Not an option. But in the options...
\(26^4\) appears only one time, whereas
\(26^3\) and \(26^5\) appear FOUR times
Get rid of \(26^4\)

\(26^3 + 26^4 + 26^5 =\)

\((26^3 + 26^4) + 26^5 =\)

\(26^3(1 + 26^1) + 26^5 =\)

\(26^3(27) + 26^5 =\)

\((27)26^3 + 26^5 =\)


Answer D

*"FCP without restriction"? For each code, choosing the first letter does not affect the second choice.
If I choose M for slot 1, I can choose M for slot 2, and M for slot 3. That is, after the choice for slot 1, there are
no restrictions on subsequent choices/slots. "FCP w/o restriction."

IF, however, the prompt said, "no repeat letters," we would have "FCP with restriction":
First choice: 26 letters. Can't repeat letters. The choice for slot 1 removed one (restricted the) choice for slot 2.
Second choice: only 25 possibilities. Third choice: 24 possibilities.
Choices for subsequent slots ARE restricted. 3-letter code using FCP with restriction: (26*25*24)

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Re: All of the bonds on a certain exchange are designated by a 3   [#permalink] 15 Apr 2018, 10:25
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