TIFR CSE 2024 | Part B | Question: 2

Question

Let $\text{S}$ be the set of all $4$ -digit numbers created using just the digits $1,2,3,4,5$ such that no two successive digits are the same. If the numbers in $\text{S}$ are arranged in ascending order, what is the $100$ th number in this sequence?

• A. $2135$
• B. $2324$
• C. $2315$
• D. $2352$
• E. $2415$

   

Answer 1

An approach of position possibilities of 4 digits can solve this efficiently.

As they have told we arrange in ascending order lets start with first position fixed as _1_ 

In the next 3 positions of 4 digits

2^nd position =  _4_ (2,3,4,5 = 4)

3^rd position = _4_(includes 1 and leaving the used digit in 2^nd position so = 4 ) 

4^th position = _4_(leaving the used digit in 3^rd position = 4)

total = 4*4*4 = 64 possibilities when 1 is fixed at first position

Now lets fix _2_ in first position then fix _1_ in second position the possibilities in

3^rd position = _4_(includes 2 and leaving the used digit in 2nd position i.e 1 so = 4 )

4^th position = _4_(leaving the used digit in 3rd position so = 4 )

so 4*4 = 16 

total 64 +16 = 80

Now to fix _2_ in first position then fix _2_ in second position is not possible as that no two successive digits are the same is the condition.

Now lets fix _2_ in first position then fix _3_ in second position the possibilities are

3rd position = _4_(includes 2 and leaving the used digit in 2nd position i.e 3 so = 4)

4th position = _4_(leaving the used digit in 3rd position so = 4)

so 4*4 = 16 

total 80 +16 = 96

next we need 4 numbers to get 100^th one so 2412, 2413, 2414, 2415  

Hence the option is E – 2415