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How many 5 letter (lower case) passwords are possible with with at least 2 'a's?
| 73 views
0
Is it 16276?
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It says 162626,

+2
Small letters total 26 alphabets

No of 5 letter passwords using two a's= 1*1*25*25*25=15625

No of 5 letter passwords using three a's= 1*1*1*25*25=625

No of 5 letter passwords using four a's=1*1*1*1*25=25

No of 5 letter passwords using five a's = 1*1*1*1*1=1

Total no of passwords having atleast two a's=15625+625+25+1=16276

(25 is for alphabets apart from a)
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using two a's =1*1*25*25*25=15625

does it cover case like two a's not together (out of 5 we selected 2 but didn't arrange )

lly for 3a's and 4 a's I think you just missed that OR may be I
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I think it's included in the cases that I have considered.

Like if you consider two a's are not together, then we will get expression as 1*25*25*25*1=15625 so ultimately the result will be same.

Please correct me if I am wrong
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is it  162626 ?

How many 5 letter (lower case) passwords are possible with with at least 2 'a's?

number of ways of filling a box with all letter = 26 and without a = 25.

solution :

way 1:

possible passwords with 2a's , 3a's, 4a's or 5a's.

way 2:

total possible passwords with all lower case alphabets - (0 a's + 1 a's).

following way 2  as its short :

possible passwords with 0 a's = 255=9765625

possible passwords with 1 a = 5C1 * 254=1953125

(no of ways a can be placed in 5 places = 5C1 and other 4 places can be filled by other 25 alphabets in 254 ways)

required possible way = 265-255 - 5* 254

=162626

if we use way 1 :

No of 5 letter passwords using two a's= 5C2 * 253=156250

No of 5 letter passwords using three a's= 5C3 * 252=6250

No of 5 letter passwords using four a's=5C4 *25=125

No of 5 letter passwords using five a's = 5C5 =1

by Boss (12.2k points)
selected
there are 26 latters

total no of password possible(containing each case)=26*26*26*26*26=11881376

no of password containing 0(zero) no of a's=25*25*25*25*25=9765625

no of password containing 1 no of a's=1*25*25*25*25*5=1953125 (we multiply by 5 because single a can occur at 5 different position )

total no of password at least two a's=total no of password - (total no of password containing 0 no of a's) - (total no of password containing  1 no of a's)=11881376-9765625-1953125=162626
by Boss (10.6k points)