A 5-digit code consists of one number digit chosen from 1, 2

Question

A 5-digit code consists of one number digit chosen from 1, 2, 3 and four letters chosen from A, B, C, D, E. If the first and last digit must be a letter digit and each digit can appear more than once in a code, how many different codes are possible?

A. 375
B. 625
C. 1,875
D. 3,750
E. 5,625

A. 375
B. 625
C. 1,875
D. 3,750
E. 5,625

Bunuel · Accepted Answer

A 5-digit code consists of one number digit chosen from 1, 2, 3 and four letters chosen from A, B, C, D, E. If the first and last digit must be a letter digit and each digit can appear more than once in a code, how many different codes are possible?

A. 375
B. 625
C. 1,875
D. 3,750
E. 5,625

Notice that each digit can appear more than once in a code.

Since there should be 4 letters in a code (X-X-X-X) and each letter can take 5 values (A, B, C, D, E) then total # of combinations of the letters only is 5*5*5*5=5^4.

Now, we are told that the first and last digit must be a letter digit, so number digit can take any of the three slots between the letters: X-X-X-X, so 3 positions and the digit itself can take 3 values (1, 2, 3).

So, total # of codes is 5^4*3*3=5,625.

Answer: E.

Similar question to practice: https://gmatclub.com/forum/a-4-letter-c ... 59065.html

Hope it helps.

KarishmaB · Answer

You are allowed duplicates. Even if A appears in the first as well as the last position, it will give you a code different from what you get when you have different letters in the first and the last position. You need to arrange the letters here. If instead you needed to just select groups, then yes, you would have worried about the effect of duplicates.

You select a letter for the first position in 5C1 ways and a letter for the last position in 5C1 ways.
Say you put the digit in the second position. You can select a digit for the second position in 3C1 ways.
You can select the letters for the third and fourth positions in 5C1 and 5C1 ways. 
Hence, you get 5*3*5*5*5 codes.
But here, we have put the digit in the second place. It could have been in the third or fourth place too. So you multiply the above given result by 3. 
Hence total number of codes = 5*3*5*5*5*3 = 5625

GyanOne · Answer

Choose one number from 3 numbers in 3C1 = 3 ways
Choose one position from the middle three for the number in 3C1 = 3 ways
The other four positions can be filled by the 5 letters in 5^4 ways.

Therefore total number of codes possible = 3*3*(5^4) = 5,625

E

SFF · Answer

For the first and last positions the letters can be chosen in 5c1 ways, but if we get say the same letter twice then we need to divide by 2! to avoid duplicates since we are looking for different codes.

For the 3 positions in between, we can choose 1 digit in 4c1 ways, and the remaining 2 digits in 5c1 and 5c1 ways. These three can rearrange themselves in 3! ways and we again divide by 2! to avoid duplicates since we can have the letters to repeat themselves. I have shown this below :-

5c1  /2! 5c1

1 2 3 4 5

Therefore total according to me would be :-

* /2! = (25*25*6*4)/4 = 3750.

Can anyone correct where i am making a mistake.

Bunuel · Answer

Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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honchos · Answer

Karishma can you tell me wher I am wrong-
1st and last places can be filled in 5 X 5 ways, another 3 this way

5C1X5C1X3C1 and they can be arranged among 3! ways = 5C1X5C1X3C1 x3! = 5x5x3x3x2

finally
5x5x5x5x3x3x2

KarishmaB · Answer

I am not sure how you have worked this out.

5*5 is fine for first and last positions - they must be letters. But you need 5 digit code so you have another 3 positions to fill

* _ _ _ *

You must use one number digit so you can select a number digit in 3 ways and the position for that number digit in 3 ways. 
Also the other two positions must be letters so they can be selected in 5*5 ways.

In all, 5*5*3*3*5*5

gmatser1 · Answer

Can someone explain why the first and last digits are 5^2 and not just 5 options?

koreye · Answer

since repetitions are allowed, there are 5 ways to choose the letters for the first slot, and five ways to choose for last slot

thats 5 x _ x _ x _ x 5

you still have to choose 1 digit and 2 letters between. For the letters, again, repetitions are allowed so each slot can be filled in 5 ways. For digit, it can only be filled in 3 ways. However, since the digit can be in any one of the 3 middle slots, so we multiply by 3 again:

5 x 3 x 5 x 5 x 5 or 
5 x 5 x 3 x 5 x 5 or 
5 x 5 x 5 x 3 x 5

so 5 x 5 x 5 x 5 x 3 x 3 = 5^4 x 3^2 = 5625.

ScottTargetTestPrep · Answer

Let’s determine the number of ways we can produce each digit.
 
If L denotes a letter digit and N denotes a number digit, the possibilities for the code are L-N-L-L-L, L-L-N-L-L, and L-L-L-N-L. Note that there are an equal number of possible codes for each of these formats, therefore we will find the number of L-N-L-L-L codes and multiply the result by three.
 
Since the first digit must be a letter, we have 5 options for the first digit. Since the second digit is a number, there are 3 options for the second digit. For the third, fourth, and fifth digits, we have 5 options each. In total, there are 5 x 3 x 5 x 5 x 5 = 1875 L-N-L-L-L codes. Since the total number of codes is three times that, there are 1875 x 3 = 5625 possible codes. 
 
Answer E

D4kshGargas · Answer

Hi, I am having a hard time understanding the statement "If the first and last digit must be a letter digit"
Doesn't that mean the first and last digit must be a  ? Like A1 or B1, something like that? 
so the code will be 
A1 _ _ _ B1, something like that

I know I'm wrong but can you please clarify? 
Thanks!

Bunuel · Answer

The first and the last symbols must be letters. For example,  A AB2 B ,  C 3AC C , ...

Crytiocanalyst · Answer

To occupy the first position we have to choose one number from 3 numbers in 3C1 = 3 ways

Another one position is accompanied by a number in 3C1 = 3 ways

The remaiing four positions can be filled in 5^4 ways.

Therefore the total no of codes that can be made = 3*3*(5^4) = 5,625

Hence IMO E

Kinshook · Answer

Given: A 5-digit code consists of one number digit chosen from 1, 2, 3 and four letters chosen from A, B, C, D, E. 
Asked: If the first and last digit must be a letter digit and each digit can appear more than once in a code, how many different codes are possible?

Number of ways to chose first and last digits = 5^2
Number of ways to chose remaining digits = 3C1*3*5^2

Number of different codes possible = 3C1*3*5^4 = 3^2*5^4 = 9*625 = 5625

IMO E

sarthak1701 · Answer

First and last can be chosen in 5 ways. Let's put 5 in both positions

5 _ _ _ 5

Now, the code must contain 1 digit and 4 letters and we have 3 digits and 5 letters to choose from.

We can choose a digit in 3 ways and a letter in 5 ways, which gives us 3*5*5 possible codes BUT we also need to adjust for the arrangement within these 3 letters, DLL (Digit - Letter - Letter) can be arranged in 3!/2! ways = 3 ways. Therefore, the middle 3 letters can be chosen in 3*5*5*3 ways

Plug these values in the positional text above and we get = 5*3*5*5*3*5 = 5625

Kinshook · Answer

A 5-digit code consists of one number digit chosen from 1, 2, 3 and four letters chosen from A, B, C, D, E.

If the first and last digit must be a letter digit and each digit can appear more than once in a code, how many different codes are possible?

Number of choices to select 4 letter digits = 5^4

L_L_L_L

Since first and last digits must be a letter digit, there are 3 spaces to choice for number digit and there are 3 choices to chose number digits

Number of different codes possible = 5^4*3*3 = 5625

IMO E

SwethaReddyL · Answer

hi  Bunuel  why are we not dividing the duplicates if reptation is allowed?

Bunuel · Answer

We do not divide because the method already counts each code exactly once.

Here, the positions are distinct. For example, AAB1C is formed by choosing A for the first slot, A for the second, B for the third, 1 for the number slot, and C for the last slot. That code is counted once, not multiple times. Repetition allowed only means a symbol may be used again in another slot.

Adit_ · Answer

Seems pretty straightforward.

First and last has 5 choices each so multiply them.
You have 3 choices in between out of which 1 is a number digit:
Choosing one of the 3 spots * 3 choices = 3c1*3 = 9
Remaining two spots again 5 choices of letters.
Thus we have 5*5*5*5*9 = 9*5^4 = 9*625 = 5625.

Answer:  Option E

A 5-digit code consists of one number digit chosen from 1, 2

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A 5-digit code consists of one number digit chosen from 1, 2

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