ISI Kolkata M.Tech CSE Interview Experience (2025)

Question

Name: Param Modi
Gate Score: 711
Gate Rank: 1121

I was shortlisted for the M.Tech CSE interview at ISI Kolkata directly based on my GATE score, without having to appear for their written entrance exam.

Interview Date: 25th April 2025
Duration: 35–45 minutes
Panel: 6 Professors

They first asked me about my favourite subjects in Mathematics and Computer Science.
I mentioned:

• Mathematics: Probability, Discrete Mathematics

• Technical: Data Structures & Algorithms (DSA), Theory of Computation (TOC)

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Mathematics Questions

Q1. What is the expectation of getting two consecutive heads with a fair coin?
✅ Solved
They asked me my approach after solving.

Q2. Using an unfair coin (with given probabilities), what is the probability of getting at least one head in two tosses?
✅ Solved
They asked me my approach after solving.

Q3. How many N-digit numbers can be formed using digits 0–9 such that no two consecutive digits are numerically consecutive? (0 cannot be the first digit)
✅ Solved
They asked me my approach after solving.

Q4. How many N-digit numbers are possible such that no three consecutive digits are the same? (two consecutive same digits are allowed)
❌ Not solved

Q5. Given: 34x = 43y. Prove that x + y is a composite number.
❌ Not solved

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Technical Questions

Q6. Is the language over {0,1} that does not contain three consecutive 1s regular?
✅ Solved

Q7. How?
✅ Solved — I said I could build a DFA.

Q8. Construct the DFA
✅ Solved

Q9. Write the regular expression for the same language
❌ Not solved

Q10. Write the regular expression for the language that contains '111' as a substring
✅ Solved

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Result: Offered admission in the first round to the M.Tech CSE program at ISI Kolkata.

Comments

Congratulations 🎉

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Given: 34x = 43y. Prove that x + y is a composite number.

Solution:- 

33x + x = 44y - y

x + y = 44y - 33x

x + y = 11(4x - 3y)

Since its multiple of 11 hence its composite.

Congratulations, wishing u good luck ahead. Are there more rounds left?

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Very Easy and understandable solution. thank you for solving..

And Yes, there might be more rounds left as they have kept few people on waitlist.

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Buddy may to chal rha hai or 25 may 2025 to  abhi aayi bhi nhi hai

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congrats brother, what is your category(gen/obc?) and till what gate score did they call for direct interiews for gen/obc?

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Mine is OBC.
for GEN , They called till 750 Gate score
for OBC, They called till 650 Gate score

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brother how did u solve the question 3. How many N-digit numbers can be formed using digits 0–9 such that no two consecutive digits are numerically consecutive? (0 cannot be the first digit)

please recall it . im not able to do this and its frustrating since few days
My guess is this can be solved using inclusion -exclusion principle by

 For N=2 digits  by making cases
if first digit is 1 then we have 9 ways for 2nd digit( other than 2- as it is numerically consecutive)
if first is 2 then again 9 ways for 2nd digit (other than 3)..
if first is 8 then again 9 ways for 2nd digit
further if first is 9 then we have 10 ways ( including zero as its not numerically consecutive to 9) so totally 8*9+10 = 82 -2digit numbers .

For N=3 digits then
take first 2 digits it will be 82 such no. that satisfies the condition and then
take 2nd and 3rd digit again 82 such no.'s adding these both then
we need to  subtract no.'s where all  3 digits are numerically consecutive  that are 7 so this way we get
total = 82*2 - 7 as 3 digit numbers  .

for N=4
take first 2 digits then
2nd and 3rd digits and then
3rd and 4th digits and solve by Inc-exc principle (by adding each of these 2 digit case and then subtracting 3digits case and then adding 4 digits case )

im not sure if this is effective solution . please put your thoughts about it !!

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Hey, Sorry for replying late. 
You can check this question which is similar to question 3 above. 
I have answered similar kind of question here : Click Here...

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How many N-digit numbers are possible such that no three consecutive digits are the same? (two consecutive same digits are allowed)? (0 cannot be the first digit)

 

Let R(n) be a function that computes the result for a given n.
R(1) = 9
R(2) = 9*10 = 90

R(n-1) is count of numbers with n-1 digits. We can add one more digit that is not same as the n-1th digit. There are 9 possible choices.We get 9*R(n-1) strings of length n.

Now which strings are missing from the count?
The strings ending with two consecutive digits like 8911, 123455.

If the string ends with 2 consecutive digits, n-1th digit and n-2th digit cannot be same. So we find R(n-2) and add one more digit that is not same as the last digit of the strings counted by R(n-2). There are 9 choices. Also (n-1)th digit will be same as nth digit. So count of such strings are 9*R(n-2). 

R(n) = 9*R(n-1)+9*R(n-2)

Solving this recurrence will give a generalized formula for R(n).