In how many ways can 21 identical Red balls and 19 identical Blue ball

Question

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?

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

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

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

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

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

BrentGMATPrepNow · Answer

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. 
So the question becomes, "  In how many different ways can we select 19 spaces from 22 spaces?  "
Since the order in which we select the 19 spaces does not matter, we can use combinations.

We can select 19 spaces from 22 spaces in 22C19 ways. 
22C19 = 22!/(19!)(3!)

Answer: A

dave13 · Answer

BrentGMATPrepNow i applied concept putting all blue together and then subtracting from total number of arrangements but got stuck at getting final answer ..

1.) so total number of combinations: \(\frac{40!}{21!19! } \)

2.) putting all blue ones together: \(\frac{22!}{19!}\)

\(\frac{40!}{21!19! } \) - \(\frac{22!}{19!}\) = here i got stuck how to get final answer

help is appreciated

BrentGMATPrepNow · Answer

How did you calculate part 2 (putting all blue ones together)?

Keep in mind that there are many different ways to break the restriction that no two blue balls are together. 
For example, there could just be 2 blue balls together, or 3 blue balls together, or two pairs of 2 blue balls together, or three pairs, or three pairs of 2 and one set of 3 in a row, etc....
I don't believe 22!/19! covers all possible outcomes.

dave13 · Answer

BrentGMATPrepNow yes i calculated all blue ones by merging them all into one blue. i though by merging all blue ones ii would consider all possible options you mentioned... can you please give an example of problem where i would be able to use the concept i used

Regor60 · Answer

The 19 blue balls are separated by 18 spaces.

Satisfy the condition that they not be contiguous by placing 1 red ball in each space.

This leaves 3 red balls remaining to be placed. There are 18 spaces plus one on either end, total of 20 spaces to place remaining 3.

These 20 space are separated by 19 dividers.

The 19 dividers + 3 red balls are 22 objects that can be arranged 22!.

However, to eliminate duplicates, divide by 19! and 3! = 22!/3!19!

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sritamasia · Answer

First, arrange 21 Red balls in a row. This can be done in 1 way (since they are identical). 
Now there are 22 places for the 19 blue balls and so the places can be filled in 
22C19 ways =  \frac{22!}{3!19!}

Kinshook · Answer

Asked: 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?

_R_R_R_R_R_R_R_R_R_R_R_R_R_R_R_R_R_R_R_R_R_

Total number of ways to arrange 21 identical Red balls and 19 identical Blue balls in a row so that no 2 Blue balls are together = 22C19 = 22!/3!19!

IMO A

BrentGMATPrepNow · Answer

Here's one such question:  In how many ways can 21 identical Red balls and 19 identical Blue balls be arranged in a row so that all 19 blue balls are not together?

bumpbot · Answer

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In how many ways can 21 identical Red balls and 19 identical Blue ball

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In how many ways can 21 identical Red balls and 19 identical Blue ball

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