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A 5-digit code consists of one number digit chosen from 1, 2 [#permalink]
08 May 2012, 22:30

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00:00

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Difficulty:

55% (hard)

Question Stats:

53% (03:08) correct
48% (02:05) wrong based on 160 sessions

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?

Re: Maths Question on Combinations [#permalink]
08 May 2012, 22:46

3

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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

Re: Maths Question on Combinations [#permalink]
09 May 2012, 00:14

GyanOne wrote:

Choose one number from 3 numbers in 3C1 = 3 ways Choose four letters from 5 letters in 5C4 = 5 ways Choose one position from the middle three for the number in 3C1 = 3 ways The other four positions can be filled by the 4 letters in 4^4 ways.

Therefore total number of codes possible = 3*5*3*(4^4) = 45*16*16 = 11,520

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 :-

Re: A 5-digit code consists of one number digit chosen from 1, 2 [#permalink]
09 May 2012, 00:23

5

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Expert's post

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samarthgupta wrote:

A friend asked me this question recently and I wasn't able to get the official answer for this question. I am not sure about the source or the difficulty level. The questions is as follows :-

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).

Re: Maths Question on Combinations [#permalink]
09 May 2012, 08:55

3

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Expert's post

samarthgupta wrote:

GyanOne wrote:

Choose one number from 3 numbers in 3C1 = 3 ways Choose four letters from 5 letters in 5C4 = 5 ways Choose one position from the middle three for the number in 3C1 = 3 ways The other four positions can be filled by the 4 letters in 4^4 ways.

Therefore total number of codes possible = 3*5*3*(4^4) = 45*16*16 = 11,520

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 :-

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 _________________

Re: Maths Question on Combinations [#permalink]
18 Sep 2013, 04:56

VeritasPrepKarishma wrote:

samarthgupta wrote:

GyanOne wrote:

Choose one number from 3 numbers in 3C1 = 3 ways Choose four letters from 5 letters in 5C4 = 5 ways Choose one position from the middle three for the number in 3C1 = 3 ways The other four positions can be filled by the 4 letters in 4^4 ways.

Therefore total number of codes possible = 3*5*3*(4^4) = 45*16*16 = 11,520

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 :-

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

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

Re: Maths Question on Combinations [#permalink]
18 Sep 2013, 06:02

Expert's post

honchos wrote:

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

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.