GRE® Quant Practice | Permutation Combination

GRE Sample Questions | Reordering of Letters - With Constraints

This GRE quant practice question is a permutation combination problem solving question. A classic example of reordering letters of a word with an interesting constraint. Level of Difficulty: HARD

Question 5: In how many rearrangements of the letters of the word SCINTILLATING will no two 'I's appear together?

  1. 11C3 * 13!
  2. \\frac{10!}{2! × 2! × 2!})
  3. 11C3 × 3! × 10!
  4. 11C3\\frac{10!}{2! × 2! × 2!})
  5. \\frac{11!}{2! × 2! × 2!})

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Explanatory Answer | GRE Permutation Combination Practice Question 5

Step 1 of solving this GRE Combination 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 Ns, 2 Ts and once each of S, C, A, and 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: _ S _ C _ N _ T _ L _ L _ A _ T _ N _ 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 11C3 ways.

Step 2 of solving this GRE Combination 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 = \\frac{10!}{2! × 2! × 2!})

Therefore, the total number of ways the letters of the word SCINTILLATING can be reordered,
where no two 'I's appear together is 11C3 × \\frac{10!}{2! × 2! × 2!})

Choice B 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?

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