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Help Me out with this Combinotrics prob. from GMATPREP

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by DREAM GMAT » Sat Oct 10, 2009 9:07 pm
Each stack is designated with a 1, 2, and 3 letter code where each letter is selected from the 26 letters of the alphabet. If the letters may be repeated and if the same letters used in a different order constitutes a different code, how many different stocks is it possible to uniquely designate with these codes?
a. 2951
b. 8125
c. 15600
d. 16302
e. 18278

I Guessed the answer which was correct, but need the explanation for the same.

Thanks
MANDEEP
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by xcusemeplz2009 » Sun Oct 11, 2009 2:34 am
IMO E

26+26*26+26*26*26
It does not matter how many times you get knocked down , but how many times you get up
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by rdchandvadkar » Fri Feb 12, 2010 1:15 pm
could some one explain this one in a lil detail..

this is what i did.

for 1 alphabet code- 26 ways.
for 2 alphabet code- 26P2= 26*25
for 3 alphabet code- 26P3= 26*25*24

and then add them together....

whats is wrong... im missing something here.. pls explain from the start.. permutation and combination is a major bother for me.

thanks..
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by rdchandvadkar » Fri Feb 12, 2010 1:26 pm
i think i got the logic...

for 1 alphacode: 26 ways..
for 2 alphacode: the 1st alpha can be selected from 26 available; the 2nd alpha can be selected from 26 available alpha(as the same alpha can be used)
similarly for 3 alphacode.

therefore 26+26^2+26^3


hope this is the right logic i have used this time around.
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by thephoenix » Fri Feb 12, 2010 7:23 pm
rdchandvadkar wrote:i think i got the logic...

for 1 alphacode: 26 ways..
for 2 alphacode: the 1st alpha can be selected from 26 available; the 2nd alpha can be selected from 26 available alpha(as the same alpha can be used)
similarly for 3 alphacode.

therefore 26+26^2+26^3


hope this is the right logic i have used this time around.
yes this the right approach for this one since the question says repetetion is allowed for each pace we hace 26 choices

had there been no repetetion.....the ans wud have been

26c1+26c2+26C3=2951
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