in Combinatory retagged by
224 views
5 votes
5 votes

In how many rearrangements of the letters of the word SCINTILLATING will no two 'I's appear together?

  1. ${ }^{11} \mathrm{C}_{3} * 13!$
  2. $\frac{10 !}{2 ! * 2 ! * 2 !}$
  3. ${ }^{11} \mathrm{C}_{3} * 3 ! * 10 !$
  4. ${ }^{11} \mathrm{C}_{3} \frac{10 !}{2 ! * 2 ! * 2 !}$
in Combinatory retagged by
224 views

1 Answer

6 votes
6 votes

Step $1$ of solving this question: Segregate the word into 'I's and letters that are not 'I's:

SCINTILLATING is a $13$ letter word comprising $3$ Is, $2$ Ls, $2 \mathrm{Ns}, 2$ Ts and once each of $\mathrm{S}, \mathrm{C}, \mathrm{A}$, and $\mathrm{G}$.

The condition given is that no two 'I's should appear together. The same condition can be reworded as "there should be at least one letter in between any two 'I's The way to achieve this condition is to place the remaining $10$ letters in such a way that there is exactly one gap between any two of these letters.

By this action we are ensuring that not more than one 'I' can be placed in the gap - making sure that no two 'I's appear together.
It will appear like this: $$_{-}\mathrm{S}_{-} \mathrm{C}_{-}\mathrm{N}_{-}\mathrm{T}_{-}\mathrm{L}_{-} \mathrm{L}_{-} \mathrm{A}_{-} \mathrm{T}_{-}\mathrm{N}_{-} \mathrm{G}_{-}$$ There are a total of $11$ places where the 'I's can be placed including the one before the first letter and the one after the last letter - and no two 'I's will appear together.

We have $3$ 'I's. So, we need to choose $3$ out of $11$ places to place the 'I's. The $3$ 'I's can be placed in $11$ places in $11$ choose $3$ ways or $11 \mathrm{C} 3$ ways.
Step $2$ of solving this question: Reorder the remaining letters:
Now the remaining $10$ letters have to be reordered.
These $10$ letters comprise $2$ 'L's, $2$ 'N's and $2$ 'T's.
The number of ways to reorder these $10$ letters $=$ $=10 ! /(2 ! \times 2 ! \times 2 !)$

Therefore, the total number of ways the letters of the word SCINTILLATING can be reordered,
where no two 'I's appear together is $11 \mathrm{C} 3 \times 10 ! /(2 ! \times 2 ! \times 2 !)$
Choice D is the correct answer.

Similar Questions:
The following questions are variants of the same concept.

  1. In how many ways can $3$ boys and $3$ girls be made to stand in a line such that no two boys stand together?
  2. In how many ways can $10$ balls, $4$ which are black and identical and $6$ of which are white and identical be arranged in a line such that no two black balls are placed next to each other?


Answer these questions in the comments.

edited by

4 Comments

@Vaibhav04 Excellent.

1
1
After choosing 3 Iā€™s from 11 gaps it is choosen 11c3 ways and after that why not it be arranging 3 factorial ways so the atlast ans will be 11c3*3 factorial for no two i s appear together?
0
0

@Amlan Kumar Majumdar We are not multiplying by 3! because all the 3 I's are identical. Although 3 items can be arranged in 3! ways, since all these are identical, we divide by 3! (refer to Division principle) and that will be 11C3 * 3!/3! = 11C3

All the 3! = 6 ways will be identical

ie.  i1 i2 i3 = i1 i3 i2 = i2 i1 i3 = i3 i2 i1= i2 i3 i1 = i3 i1 i2 = i i i 

 

    

 

 

0
0
Answer:

Related questions