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A certain identification code is a list of five symbols: S1S

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A certain identification code is a list of five symbols: S1S [#permalink] New post 25 Feb 2017, 08:16
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88% (00:39) correct 11% (01:49) wrong based on 94 sessions


A certain identification code is a list of five symbols: S1,S2,D1,D2,D3. Each of the first 2 symbols must be one of the 26 letters of the English alphabet, and each of the last 3 symbols must be one of the 10 digits. (Repeated letters and digits are allowed.) What is the total number of different identification codes?


A) 1,757,600

B) 676,000

C) 468,000

D) 1,676

E) 82
[Reveal] Spoiler: OA

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Re: A certain identification code is a list of five symbols: S1S [#permalink] New post 04 Mar 2017, 07:40
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Explanation

Now here we have 5 places to fill S1, S2, D1, D2 and D3 respectively. Repetition is allowed

S1: can take vale of 26 alphabets (26 options)
S2: can take vale of 26 alphabets (26 options)
D1: can take vale of 10 digits (10 options)
D2: can take vale of 10 digits (10 options)
D3: can take vale of 10 digits (10 options)

Total no of possibilities = \(S1 \times S2 \times D1 \times D2 \times D3\)= \(26 \times 26 \times 10 \times 10 \times 10\) = 676,000.

Hence B is the correct option.
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Re: A certain identification code is a list of five symbols: S1S [#permalink] New post 07 Sep 2018, 17:57
Hi Sandy, can you explain why we couldn't divide the product of the first two letters by two? Identical letters would count two identical outcomes right? (e.g., AA)
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Re: A certain identification code is a list of five symbols: S1S [#permalink] New post 08 Sep 2018, 12:42
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jasonchung55 wrote:
Hi Sandy, can you explain why we couldn't divide the product of the first two letters by two? Identical letters would count two identical outcomes right? (e.g., AA)


Look at this:

Group 1



AA, AB,............. AZ ----- 26 items
BA, BB, .............BZ ----- 26 items


ZA, ZB, ............. ZZ ---- 26 items

Here total number of combinations possible is \(26*26\)

Similarly, we have 3 digits


Group 2



000, 001, ...... 009 ---- 10 items
010, 011, .......019 ---- 10 items

.
.
.
.
990, 991,992 .....999 ---- 10 items

Here total number of combinations is \(10*10*10\)

Each item in group 1 has the complete group 2 associated with it.


So

AA and

000, 001, ...... 009 ---- 10 items
010, 011, .......019 ---- 10 items

.
.
.
.
990, 991,992 .....999 ---- 10 items

is one part of the set of solutions. Also,

AB and

000, 001, ...... 009 ---- 10 items
010, 011, .......019 ---- 10 items

.
.
.
.
990, 991,992 .....999 ---- 10 items

So on. Thus the total number of combinations is \(26*26*10*10*10\)


If you look at this

AA, AB,............. AZ
BA, BB, .............BZ
CA, CB, CC ..........CZ

ZA, ZB, .............ZZ

Here AA is not counted twice. Hence the answer is correct!
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Re: A certain identification code is a list of five symbols: S1S [#permalink] New post 13 Sep 2018, 12:25
Would really like it if the options lit up as Green or Red :D

Ans = 26*26*10*10*10 = 676,000
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Re: A certain identification code is a list of five symbols: S1S [#permalink] New post 04 Nov 2018, 12:40
I don't understand how AA123 is counted twice. Please explain.
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Re: A certain identification code is a list of five symbols: S1S [#permalink] New post 04 Nov 2018, 14:06
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I have modified the explanation above thanks for pointing it out!
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Re: A certain identification code is a list of five symbols: S1S [#permalink] New post 20 Dec 2018, 03:05
Sandy here it is given how many diffrent?
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Re: A certain identification code is a list of five symbols: S1S [#permalink] New post 20 Dec 2018, 04:04
if repetitions were not allowed how would the answer be different?




sandy wrote:
Explanation

Now here we have 5 places to fill S1, S2, D1, D2 and D3 respectively. Repetition is allowed

S1: can take vale of 26 alphabets (26 options)
S2: can take vale of 26 alphabets (26 options)
D1: can take vale of 10 digits (10 options)
D2: can take vale of 10 digits (10 options)
D3: can take vale of 10 digits (10 options)

Total no of possibilities = \(S1 \times S2 \times D1 \times D2 \times D3\)= \(26 \times 26 \times 10 \times 10 \times 10\) = 676,000.

Hence B is the correct option.
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Re: A certain identification code is a list of five symbols: S1S [#permalink] New post 22 Dec 2018, 16:09
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If repetitions were not allowed:

Total no of possibilities = \(S1 \times S2 \times D1 \times D2 \times D3\)= \(26 \times 25 \times 10 \times 9 \times 8\)
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Re: A certain identification code is a list of five symbols: S1S [#permalink] New post 15 Sep 2019, 06:53
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Each of the first 2 symbols must be one of the 26 letters of the English alphabet, and each of the last 3 symbols must be one of the 10 digits. (Repeated letters and digits are allowed.) What is the total number of different identification codes?

26x26x10x10x10... repeat is allowed

Calculate 26x26 and from there don't even do the math on 10x10x10 since you know you are just adding 0s to the end of the first calculation

B
Re: A certain identification code is a list of five symbols: S1S   [#permalink] 15 Sep 2019, 06:53
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