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In how many ways 21 red balls and 19 blue balls can be arranged in a row so that no two blue balls are together?
- a)1540
- b)1520
- c)1560
- d)None
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
In how many ways 21 red balls and 19 blue balls can be arranged in a r...
As there are 21 red balls and 19 blue balls.
So there can be 22 different positions for blue balls ( because we can't arrange two blue balls together).
Applying concept,
Total ways = 22C19
= 22!/19!(22-19)!
= 22!/19!*3!
= 22*21*20*19!/19!*3!
= 22*7*10
= 1540.
Most Upvoted Answer
In how many ways 21 red balls and 19 blue balls can be arranged in a r...
Given:
Total number of red balls = 21
Total number of blue balls = 19
To find:
Number of ways in which the balls can be arranged in a row so that no two blue balls are together
Solution:
Let us first arrange all the red balls in a row. This can be done in 21! ways.
Now, we need to arrange the blue balls in such a way that no two blue balls are together.
To do this, we can first arrange all the blue balls in a row. This can be done in 19! ways.
We now have 20 positions (represented by the gaps between the red balls) in which we can place the blue balls in such a way that no two blue balls are together.
Out of these 20 positions, we need to select 19 positions for the blue balls.
This can be done in 20C19 ways = 20 ways.
Now, we can arrange the blue balls in the selected positions in 19! ways.
Therefore, the total number of ways in which the balls can be arranged in a row so that no two blue balls are together = 21! x 20 x 19! = 1540 x 10^6
Hence, option A is the correct answer.
Total number of red balls = 21
Total number of blue balls = 19
To find:
Number of ways in which the balls can be arranged in a row so that no two blue balls are together
Solution:
Let us first arrange all the red balls in a row. This can be done in 21! ways.
Now, we need to arrange the blue balls in such a way that no two blue balls are together.
To do this, we can first arrange all the blue balls in a row. This can be done in 19! ways.
We now have 20 positions (represented by the gaps between the red balls) in which we can place the blue balls in such a way that no two blue balls are together.
Out of these 20 positions, we need to select 19 positions for the blue balls.
This can be done in 20C19 ways = 20 ways.
Now, we can arrange the blue balls in the selected positions in 19! ways.
Therefore, the total number of ways in which the balls can be arranged in a row so that no two blue balls are together = 21! x 20 x 19! = 1540 x 10^6
Hence, option A is the correct answer.
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Community Answer
In how many ways 21 red balls and 19 blue balls can be arranged in a r...
21 red ball can be arranged 21 at a time
and for 19 blue ball it can be arranged 22 at a time
but as this question is not ask for different no.
so this can be arranged as 22C19 = 22!/19!*3!= 22*21*20/3*2 = 1540
so , the correct option is (a )
and for 19 blue ball it can be arranged 22 at a time
but as this question is not ask for different no.
so this can be arranged as 22C19 = 22!/19!*3!= 22*21*20/3*2 = 1540
so , the correct option is (a )
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In how many ways 21 red balls and 19 blue balls can be arranged in a row so that no two blue balls are together?a)1540b)1520c)1560d)NoneCorrect answer is option 'A'. Can you explain this answer?
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