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The number of ways in which 20 different pearls of two colours can be set alternately on a necklace, there being 10 pearls of each colour, is
  • a)
    9!x10!
  • b)
    5(9!)2
  • c)
    (9!)2
  • d)
    none of these
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
The number of ways in which 20 different pearls of two colours can be ...
Ten pearls of one colour can be arranged in 1/2 x (10 - 1)! ways. The number of arrangements of 10 pearls of the other colour in 10 places between the pearls of the first colour = 10!
∴ the required number of ways = 1/2 x 9! x 10!
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Most Upvoted Answer
The number of ways in which 20 different pearls of two colours can be ...
Ten pearls of one colour can be arranged in 1/2 x (10 - 1)! ways. The number of arrangements of 10 pearls of the other colour in 10 places between the pearls of the first colour = 10!
∴ the required number of ways = 1/2 x 9! x 10!
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Community Answer
The number of ways in which 20 different pearls of two colours can be ...
Understanding the Problem
To determine the number of ways to arrange 20 pearls of two colors (10 pearls each), we need to consider the constraints of alternating colors on a necklace.
Arrangement Constraints
- The pearls must be arranged alternately, meaning the arrangement must start and end with different colors.
- This implies that one color must occupy the odd positions and the other color the even positions.
Choosing the Arrangement
1. Select the First Color: We can choose either color to start the arrangement. Let's assume we start with Color A.
2. Arrangement on the Necklace:
- For the odd positions (1st, 3rd, 5th, ..., 19th), we will place 10 pearls of Color A.
- For the even positions (2nd, 4th, 6th, ..., 20th), we will place 10 pearls of Color B.
Calculating the Combinations
- Arranging Color A: The 10 pearls of Color A can be arranged among themselves in 10! ways.
- Arranging Color B: Similarly, the 10 pearls of Color B can also be arranged in 10! ways.
Considering Rotational Symmetry
Since the arrangement is circular (necklace), we need to account for the fact that rotations of the same arrangement are considered identical. This requires us to fix one position, reducing the arrangements by a factor of 10 (the number of positions).
Thus, the formula becomes:
Total arrangements = (10! * 10!) / 10 = (10! * 10!) / (10) = (9!) * (10!)
Final Calculation
Given the options, the correct answer corresponds to option 'B':
Total arrangements = 5(9!)².
Thus, the explanation validates that the total number of ways to arrange the pearls alternately on a necklace is indeed option 'B.'
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The number of ways in which 20 different pearls of two colours can be set alternately on a necklace, there being 10 pearls of each colour, isa)9!x10!b)5(9!)2c)(9!)2d)none of theseCorrect answer is option 'B'. Can you explain this answer?
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