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
Help Me out with this Combinotrics prob. from GMATPREP
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IMO E
26+26*26+26*26*26
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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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..
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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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.
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.
- thephoenix
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yes this the right approach for this one since the question says repetetion is allowed for each pace we hace 26 choicesrdchandvadkar 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.
had there been no repetetion.....the ans wud have been
26c1+26c2+26C3=2951