Formulas: P(n, r)- C(n,r)- 3. (4 pts ea must be letters from the standard Englis
ID: 3148743 • Letter: F
Question
Formulas: P(n, r)- C(n,r)- 3. (4 pts ea must be letters from the standard English alphabet (not case-sensitive) and the last 3 characters are digits from 0 to 9. How many possible passwords are there given each set of restrictions. (a) If characters cannot be repeated? ch) You are creating a password containing six characters. The first 3 characters (b) That do not contain an even number if characters can be repeated? (c) That contain exactly one vowel if characters can be repeated (d) That contain exactly one vowel and exactly one even number if characters cannot be repeated? (e) That contain either the letter S or the digit 4 if characters cannot be repeated?Explanation / Answer
We have to creat a password containg 6 characters out of which first 3 must be letters and last 3 must be digits.
a) 1. First character may be any alphabet out of 26 alphabets.
2. Second character may be any alphabet out of remaining 25 alphabets(as repeatation is not allowed).
3. Third character may be any alphabet out of remaining 24 alphabets(as repeatation is not allowed).
4. Fourth character may be any digit out of 10 digits.
5. Fifth character may be any digit out of 9 digits(as repeatation is not allowed).
6. Sixth character may be any digit out of 8 digits(as repeatation is not allowed).
Hence,Total No of possible password = 26*25*24*10*9*8 = 11232000.
b) 1. First character may be any alphabet out of 26 alphabets.
2. Second character may be any alphabet out of remaining 26 alphabets(as repeatation is allowed).
3. Third character may be any alphabet out of remaining 26 alphabets(as repeatation is allowed).
4. Fourth character may be any digit out of 5 digits (no even numbers e.g.1,3,5,7,9).
5. Fifth character may be any digit out of 5 digits(as repeatation is allowed).
6. Sixth character may be any digit out of 5 digits(as repeatation is allowed).
Hence,Total No of possible password = 26*26*26*5*5*5 = 2197000.
c) Lets first character is vowel and second & third characters are consonants.However,these three are interchangable with one another.Hence,we have to multiply resultant with 3 to get desired result.
1. First character may be any alphabet out of 5 vowels(a,e,i,o,u).
2. Second character may be any alphabet out of remaining 21 consonants.
3. Third character may be any alphabet out of remaining 21 onsonants(as repeatation is allowed).
4. Fourth character may be any digit out of 10 digits.
5. Fifth character may be any digit out of 10 digits(as repeatation is allowed).
6. Sixth character may be any digit out of 10 digits(as repeatation is allowed).
Hence,Total No of possible password = 3*(5*21*21)*10*10*10 = 6615000.
d) Lets first character is vowel and second & third characters are consonants.However,these three are interchangable with one another.Hence,we have to multiply resultant with 3 to get desired result.Also,Lets fourth character is even and fifth & sixth characters are odd numbers.However,these three are interchangable with one another.Hence,we have to further multiply resultant with 3 to get desired result.
1. First character may be any alphabet out of 5 vowels(a,e,i,o,u).
2. Second character may be any alphabet out of remaining 21 consonants.
3. Third character may be any alphabet out of remaining 20 consonants(as repeatation is not allowed).
4. Fourth character may be any digit out of 5 digits (even numbers e.g. 0,2,4,6,8).
5. Fifth character may be any digit out of 5 digits (odd numbers 1,3,5,7,9).
6. Sixth character may be any digit out of 4 digits(as repeatation is not allowed).
Hence,Total No of possible password = 3*(5*21*20)*3*(5*5*4) = 1890000.
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