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# 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 [#permalink]

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08 May 2012, 22:30
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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
[Reveal] Spoiler: OA
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Re: Maths Question on Combinations [#permalink]

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08 May 2012, 22:46
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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

E
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Last edited by GyanOne on 09 May 2012, 01:58, edited 2 times in total.
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Re: Maths Question on Combinations [#permalink]

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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 :-

5c1 [(4c1*5c1*5c1)3!]/2! 5c1

1 2 3 4 5

Therefore total according to me would be :-

[(5c1*5c1)/2!]*[(4c1*5c1*5c1)3!]/2! = (25*25*6*4)/4 = 3750.

Can anyone correct where i am making a mistake.
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Re: A 5-digit code consists of one number digit chosen from 1, 2 [#permalink]

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09 May 2012, 00:23
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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).

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

Similar question to practice: a-4-letter-code-word-consists-of-letters-a-b-and-c-if-the-59065.html

Hope it helps.
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Re: Maths Question on Combinations [#permalink]

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09 May 2012, 08:55
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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 :-

5c1 [(4c1*5c1*5c1)3!]/2! 5c1

1 2 3 4 5

Therefore total according to me would be :-

[(5c1*5c1)/2!]*[(4c1*5c1*5c1)3!]/2! = (25*25*6*4)/4 = 3750.

Can anyone correct where i am making a mistake.

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
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Math Expert Joined: 02 Sep 2009 Posts: 36520 Followers: 7066 Kudos [?]: 92903 [0], given: 10528 Re: A 5-digit code consists of one number digit chosen from 1, 2 [#permalink] ### Show Tags 30 Jun 2013, 23:59 Bumping for review and further discussion*. Get a kudos point for an alternative solution! *New project from GMAT Club!!! Check HERE Theory on Combinations: math-combinatorics-87345.html DS questions on Combinations: search.php?search_id=tag&tag_id=31 PS questions on Combinations: search.php?search_id=tag&tag_id=52 Tough and tricky questions on Combinations: hardest-area-questions-probability-and-combinations-101361.html _________________ Director Status: Verbal Forum Moderator Joined: 17 Apr 2013 Posts: 635 Location: India GMAT 1: 710 Q50 V36 GMAT 2: 750 Q51 V41 GMAT 3: 790 Q51 V49 GPA: 3.3 Followers: 67 Kudos [?]: 420 [0], given: 297 Re: Maths Question on Combinations [#permalink] ### Show Tags 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 :- 5c1 [(4c1*5c1*5c1)3!]/2! 5c1 1 2 3 4 5 Therefore total according to me would be :- [(5c1*5c1)/2!]*[(4c1*5c1*5c1)3!]/2! = (25*25*6*4)/4 = 3750. Can anyone correct where i am making a mistake. 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 finally 5x5x5x5x3x3x2 _________________ Like my post Send me a Kudos It is a Good manner. My Debrief: http://gmatclub.com/forum/how-to-score-750-and-750-i-moved-from-710-to-189016.html Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7118 Location: Pune, India Followers: 2128 Kudos [?]: 13625 [0], given: 222 Re: Maths Question on Combinations [#permalink] ### Show Tags 18 Sep 2013, 06:02 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. In all, 5*5*3*3*5*5 _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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

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05 Nov 2014, 06:58
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Re: A 5-digit code consists of one number digit chosen from 1, 2 [#permalink]

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12 May 2015, 04:36
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Can someone pls tell me if my approach is correct :

We need to fill these 5 blank spaces. _ _ _ _ _
We have 5 letters of which we need to pick 4. That makes it 5P4. Now 4 of these 5 letters could repeat too. So to account for this, we need to divide 5P4 by 4!
Of the 3 numbers we need to pick 1. So 3P1.
This makes it
5P4/4! * 3P1 * 5P4/4! * 5P4/4! * 5P4/4!
or 5P4/4! * 5P4/4! * 3P1 * 5P4/4! * 5P4/4!
or 5P4/4! * 5P4/4! * 5P4/4! * 3P1 * 5P4/4!

Accounting for all the scenarios we get
5P4/4! * 3P1 * 5P4/4! * 5P4/4! * 5P4/4!
+ 5P4/4! * 5P4/4! * 3P1 * 5P4/4! * 5P4/4!
+ 5P4/4! * 5P4/4! * 5P4/4! * 3P1 * 5P4/4!

=5*3*5*5*5 + 5*5*3*5*5 + 5*5*5*3*5
=5*3*5*5*5*3
=5625.
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Re: A 5-digit code consists of one number digit chosen from 1, 2 [#permalink]

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17 Jun 2015, 17:25
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Since it will have "3*3 in the number", it will have to be divisible by 9. The only choice is E.
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Re: A 5-digit code consists of one number digit chosen from 1, 2 [#permalink]

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18 Jun 2015, 05:56
Can someone explain why the first and last digits are 5^2 and not just 5 options?
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Re: A 5-digit code consists of one number digit chosen from 1, 2 [#permalink]

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18 Nov 2015, 02:59
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gmatser1 wrote:
Can someone explain why the first and last digits are 5^2 and not just 5 options?

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.
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Re: A 5-digit code consists of one number digit chosen from 1, 2 [#permalink]

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14 Dec 2015, 14:00
Hey guys, I solved this similarly, but it took me some time (4.5mins). Can someone explain what I can/should do differently? Here are my steps below.

1. Deep breath (bc combo problems give me anxiety)

2. 5 letter word ( _ _ _ _ _ )

3. First and last choices are letters ( 5 _ _ _ 5)

4. The middle includes a number so the choices are:
5 3 5 5 5
5 5 3 5 5
5 5 5 3 5

5. If we add these up, we get:
3 * (5*3*5*5*5) = 5^4 * 3^2 = 625*9 = something slightly less than 6250... (I'm quickly running out of steam here...)

6. Choose E because it is the closest answer and I've already spent way too much time on this problem
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Re: A 5-digit code consists of one number digit chosen from 1, 2 [#permalink]

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13 Jul 2016, 20:21
_ _ _ _ _

The first and last letter must be one of the letter A,B,C,D and E.

one digit would be a number out of 1,2,and 3.

So,we have 5*3*5*5*5-Digit takes the second position-1875

If digit- takes third position we have 5*5*3*5*5-1875

If digit takes the fourth position,we have 5*5*5*3*5-1875

Re: A 5-digit code consists of one number digit chosen from 1, 2   [#permalink] 13 Jul 2016, 20:21
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