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

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sritamasia
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In how many ways can 21 identical Red balls and 19 identical Blue ball [#permalink] New post   27 Nov 2021, 14:58
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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. \(\frac{21!}{3!19!}\)
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BrentGMATPrepNow
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Re: In how many ways can 21 identical Red balls and 19 identical Blue ball [#permalink] New post   27 Nov 2021, 17:27
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sritamasia wrote:
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. \(\frac{21!}{3!19!}\)


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
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Re: In how many ways can 21 identical Red balls and 19 identical Blue ball [#permalink] New post   28 Nov 2021, 03:11
BrentGMATPrepNow wrote:
sritamasia wrote:
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. \(\frac{21!}{3!19!}\)


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


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 :grin:

help is appreciated :)
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BrentGMATPrepNow
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Re: In how many ways can 21 identical Red balls and 19 identical Blue ball [#permalink] New post   28 Nov 2021, 07:09
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dave13 wrote:

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 :grin:

help is appreciated :)


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.
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Re: In how many ways can 21 identical Red balls and 19 identical Blue ball [#permalink] New post   29 Nov 2021, 04:35
BrentGMATPrepNow wrote:
dave13 wrote:

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 :grin:

help is appreciated :)


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.



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
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Re: In how many ways can 21 identical Red balls and 19 identical Blue ball [#permalink] New post   29 Nov 2021, 06:55
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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In how many ways can 21 identical Red balls and 19 identical Blue ball [#permalink] New post   29 Nov 2021, 07:35
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 = \(\frac{22!}{3!19!}\)
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Re: In how many ways can 21 identical Red balls and 19 identical Blue ball [#permalink] New post   29 Nov 2021, 07:42
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
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Re: In how many ways can 21 identical Red balls and 19 identical Blue ball [#permalink] New post   29 Nov 2021, 07:48
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dave13 wrote:
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


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?
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Re: In how many ways can 21 identical Red balls and 19 identical Blue ball [#permalink] New post   30 Nov 2022, 00:16
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