In how many ways can 21 identical Red balls and 19 identical Blue balls be arranged in a row so that no 2 Blue balls are together?

Answer Choices

Correct answer marked below

\frac{22!}{3!19!}

Correct

\frac{2!}{19!22!}

\frac{21!}{2!19!}

\frac{19!}{3!22!}

\frac{21!}{3!19!}

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Explanation

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Official Explanation

Official

Line up the 21 red balls in a row: RRRRRRRRRRRRRRRRRRR

Place a space on either side of each red ball: _ R _ R _ R _ R _ R _ R _ R _ R _ R _ R _ R _ R _ R _ R _ R _ R _ R _ R _ R _ R _ R _

Important: Each available space is a potential location for a blue ball. This ensures that no two blue balls are together.

There are 22 spaces in total.

We'll select 19 of those 22 spaces and place a blue ball in each selected space.…

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What this question tests

What it tests

Whether you can count arrangements and selections correctly — permutations vs. combinations.

Common trap

Overcounting: treating a selection as ordered when order does not matter, or forgetting to divide by duplicates.