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Intern
Joined: 16 May 2009
Posts: 28

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30 Jul 2009, 04:09
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A 4-letter code word consists of letters A, B, and C. If the code includes all the three letters, how many such codes are possible?

A) 72
B) 48
C) 36
D) 24
E) 18

I would appreciate if you can explain your steps to answer this one......
Thanks
[Reveal] Spoiler:
C
Manager
Joined: 05 Jun 2009
Posts: 75

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30 Jul 2009, 04:49
you know it contains ABC but you don't know where. So it could be ABC_ or _ABC so ABC gives you 6 options (3!) * 2 for the two different options of _ABC or ABC_ and than for the last _ it can any of the abc so thats 3 more options so * 3

so 3! * 2 *3 = 3*2*1*2*3= 36

I hope that's right.

cheers
Intern
Joined: 16 May 2009
Posts: 28

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30 Jul 2009, 04:54
yes......

plain and simple thanks SFINER
Manager
Joined: 29 Jul 2009
Posts: 114

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30 Jul 2009, 07:37
sfeiner wrote:
you know it contains ABC but you don't know where. So it could be ABC_ or _ABC so ABC gives you 6 options (3!) * 2 for the two different options of _ABC or ABC_ and than for the last _ it can any of the abc so thats 3 more options so * 3

so 3! * 2 *3 = 3*2*1*2*3= 36

well cool explanation..but i have a doubt (I am sure it is a silly one )..in the highlighted part how could u figure out that ABC_ or _ABC is the only combination.. could it not be A_BC or AB_C or C_AB etc.
Senior Manager
Joined: 25 Jun 2009
Posts: 294

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31 Jul 2009, 01:53
Pedros wrote:
A 4-letter code word consists of letters A, B, and C. If the code includes all the three letters, how many such codes are possible?

A) 72
B) 48
C) 36
D) 24
E) 18

I would appreciate if you can explain your steps to answer this one......
Thanks
[Reveal] Spoiler:
C

Alternate approach,
Total no. of codes = 4 ! ( as we 4 positions) Lets consider there will be 2 A's then its becomes 4!/2! =12
We can have 2B's as well C so 12*3 = 36.
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