Byjus Workbook

Question

Sir I am getting answer as 25

my approach 

k+1=3

k=2

n=12

nk+1=25

@sachinmittal1

@gate_cse

Comments

The question should be properly framed, as with the current wording you can answer some large number and can argue about the validity of your claim.

I believe the question should be like "What is the smallest number of jokes does the professor need ... "

Keeping this in mind, the professor need more than $12$ unique joke triples such that he never repeats a triplet in $12$ years.

Now lets assume professor has $k$ number of jokes.

Number of triples $= {k \choose 3} = \frac{k*(k-1)*(k-2)}{3*2} \geq 12$

$\therefore k * (k-1) * (k-2) ≥ 72$

With some trial and error, we find the smallest $k$ that satisfies the above inequality.

Answer :- $6$

Note - Here, the question should also mention what exactly they mean by a triple. In the above solution, I assumed triple means a set of 3.

---

with 6 jokes professor got content for 20 years :)

Answer 1

From a set of n, we need to pick 3 jokes such that no same 3 jokes are repeated,
I.e. we have to make a combination pair of 3 jokes where number of combinations >=12

I.e. nC3 >=12

n*(n-1)*(n-2)/ 3! >=12
n*(n-1)*(n-2) >= 72

One can solve the equation at this stage to get the value of n which will be equal to the answer,
But better option is to find 3 consecutive integers which satisfy the given inequality as the multiplication result is a 2 digit integral which makes it easy to find the number.
So 6*5*4 = 120 gives n=6