14 different letters of alphabet are given, words with 6 letters are formed from these given letters. How many number of words are there which have at least one letter repeated?(a) 892742(b) 999988(c) 213216(d) 786730I had been asked this question in semester exam.My question is from Counting in section Counting of Discrete Mathematics

14 different letters of alphabet are given, words with 6 letters are f

1.	14 different letters of alphabet are given, words with 6 letters are formed from these given letters. How many number of words are there which have at least one letter repeated?(a) 892742(b) 999988(c) 213216(d) 786730I had been asked this question in semester exam.My question is from Counting in section Counting of Discrete Mathematics
Answer» Right answer is (b) 999988 Easy explanation: Number of WORDS which have at least one letter REPLACED = total number of words – total number of words in which no letter is REPEATED, => 10^6 – ^12P6 => 1000000 − 924 = 999988.

Discussion

No Comment Found

Related InterviewSolutions

Determine all possibilities for the solution set of a homogeneous system of 4 equations in 4 unknowns.(a) only one(b) finitely many or zero(c) zero(d) one or infinitely manyThe question was posed to me during an internship interview.My question is from Counting in section Counting of Discrete Mathematics
Determine all possibilities for the solution set of a homogeneous system of 6 equations in 5 unknowns.(a) only one(b) zero(c) one or infinitely many(d) finitely manyThe question was posed to me in an interview for job.This is a very interesting question from Counting in section Counting of Discrete Mathematics
In a get-together party, every person present shakes the hand of every other person. If there were 90 handshakes in all, how many persons were present at the party?(a) 15(b) 14(c) 16(d) 17The question was asked in an online interview.The origin of the question is Counting topic in section Counting of Discrete Mathematics
A group of 20 girls plucked a total of 200 oranges. How many oranges can be plucked one of them?(a) 24(b) 10(c) 32(d) 7I had been asked this question in class test.My question is from Counting topic in portion Counting of Discrete Mathematics
Determine the solution for the recurrence relation an = 6an-1−8an-2 provided initial conditions a0=3 and a1=5.(a) an = 4 * 2^n – 3^n(b) an = 3 * 7^n – 5*3^n(c) an = 5 * 7^n(d) an = 3! * 5^nThis question was addressed to me during an interview for a job.This interesting question is from Advanced Counting Techniques in portion Counting of Discrete Mathematics
Determine the value of a2 for the recurrence relation an = 17an-1 + 30n with a0=3.(a) 4387(b) 5484(c) 238(d) 1437I have been asked this question in an online quiz.I need to ask this question from Advanced Counting Techniques in division Counting of Discrete Mathematics
What is the solution to the recurrence relation an=5an-1+6an-2?(a) 2n^2(b) 6n(c) (3/2)n(d) n!*3The question was posed to me in final exam.This interesting question is from Advanced Counting Techniques in division Counting of Discrete Mathematics
Determine the solution for the recurrence relation bn=8bn-1−12bn-2 with b0=3 and b1=4.(a) 7/2*2^n−1/2*6^n(b) 2/3*7^n-5*4^n(c) 4!*6^n(d) 2/8^nThis question was addressed to me during an internship interview.My question comes from Advanced Counting Techniques in portion Counting of Discrete Mathematics
The solution to the recurrence relation an=an-1+2n, with initial term a0=2 are _________(a) 4n+7(b) 2(1+n)(c) 3n^2(d) 5*(n+1)/2I got this question in a national level competition.My enquiry is from Advanced Counting Techniques in division Counting of Discrete Mathematics
Find the value of a4 for the recurrence relation an=2an-1+3, with a0=6.(a) 320(b) 221(c) 141(d) 65The question was asked in an interview for job.This interesting question is from Advanced Counting Techniques topic in section Counting of Discrete Mathematics

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment