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Question

The number of necklaces which can be formed by selecting 4 beads out of 6 beads of different coloured glasses and 4 beads out of 5 beads of different metal, is-

  1. 6P4 × 5P4 × fraction numerator 7 factorial over denominator 2 factorial end fraction    
  2. 6C4 × 5C4 × fraction numerator 7 factorial over denominator 2 factorial end fraction    
  3. 6C4 × 5C4 × fraction numerator 8 factorial over denominator 2 factorial end fraction    
  4. 6C4 × 5C4 × 7!    

Hint:

If the necklace on the left is turned over, we obtain the arrangement on the right, i.e., anticlockwise and clockwise order of arrangement is not different.
Already, we know that the number of circular permutations when anticlockwise and clockwise order of arrangement is different as (n-1)!
Hence the number of arrangement of the necklace with beads = fraction numerator left parenthesis n minus 1 right parenthesis factorial over denominator 2 factorial end fraction

The correct answer is: 6C4 × 5C4 × fraction numerator 7 factorial over denominator 2 factorial end fraction


    In all we are using 8 beads to form a necklace,
    Number of ways necklaces can be formed by selecting 4 beads out of 6 beads of different coloured glasses  = C presuperscript 6 subscript 4
    Number of ways necklaces can be formed by selecting 4 beads out of 5 beads of different metalC presuperscript 5 subscript 4
    If the necklace on the left is turned over, we obtain the arrangement on the right, i.e., anticlockwise and clockwise order of arrangement is not different.
    Hence the number of arrangement of the necklace with beads = fraction numerator left parenthesis 8 minus 1 right parenthesis factorial over denominator 2 factorial end fraction space equals space fraction numerator 7 factorial over denominator 2 factorial end fraction
    Total number of necklaces = C presuperscript 6 subscript 4 × C presuperscript 5 subscript 4 × fraction numerator 7 factorial over denominator 2 factorial end fraction

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