Maths-
General
Easy

Question

The number of ways in which three persons can dress themselves when they have 4 shirts. 5 pants and 6 hats between them, is-

  1. 4C3 × 5C3 ×6C3    
  2. 4P3× 5 P3 × 6 P3    
  3. fraction numerator 15 factorial over denominator 4 factorial 5 factorial 6 factorial end fraction    
  4. fraction numerator 15 factorial over denominator left parenthesis 3 factorial right parenthesis to the power of 3 end exponent end fraction    

hintHint:

Here, we can use the formula
C presuperscript n subscript r space cross times r factorial equals space fraction numerator n factorial over denominator left parenthesis n minus r factorial right parenthesis end fraction space space equals P presuperscript n subscript r

The correct answer is: 4P3× 5 P3 × 6 P3


    Given that, 
    Shirts = 4 ,    Pants = 5  and  Hats = 6, which are distributed among 3 men.
    For shirts, Number of ways they can wear = C presuperscript 4 subscript 3 space end subscript cross times 3 factorial
    For pants, Number of ways they can wear = C presuperscript 5 subscript 3 cross times 3 factorial
    For hats, number of ways they can wear = C presuperscript 6 subscript 3 cross times 3 factorial
    Total number of ways they can wear 
    rightwards double arrow left parenthesis C presuperscript 4 subscript 3 cross times 3 factorial right parenthesis cross times left parenthesis C presuperscript 5 subscript 3 cross times 3 factorial right parenthesis cross times left parenthesis C presuperscript 6 subscript 3 cross times 3 factorial right parenthesis

rightwards double arrow fraction numerator 4 factorial over denominator 3 factorial thin space 1 factorial end fraction cross times 3 factorial cross times thin space fraction numerator 5 factorial over denominator 3 factorial thin space 2 factorial end fraction cross times 3 factorial cross times thin space fraction numerator 6 factorial over denominator 3 factorial thin space 3 factorial end fraction cross times 3 factorial

rightwards double arrow fraction numerator 4 factorial over denominator 1 factorial end fraction cross times fraction numerator 5 factorial over denominator 2 factorial end fraction cross times fraction numerator 6 factorial over denominator 3 factorial end fraction space left parenthesis C presuperscript n subscript r space equals space fraction numerator n factorial over denominator left parenthesis n minus r factorial right parenthesis end fraction space space equals P presuperscript n subscript r right parenthesis

rightwards double arrow P presuperscript 4 subscript 3 cross times P presuperscript 5 subscript 3 cross times P presuperscript 6 subscript 3

    Thus, the number of ways in which three persons can dress themselves when they have 4 shirts. 5 pants and 6 hats between them, is P presuperscript 4 subscript 3 cross times P presuperscript 5 subscript 3 cross times P presuperscript 6 subscript 3.

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