Maths-
General
Easy

Question

Eight chairs are numbered from 1 to 8. Two women and three men wish to occupy one chair each. First women choose the chairs from amongst the chairs marked 1 to 4; and then the men select the chairs from the remaining. The number of possible arrangements is-

  1. 6C3 × 4C2    
  2. 4P3 × 4P3    
  3. 4C2 × 4P3    
  4. None of these    

hintHint:

We will first start by using the method of selecting r objects out of n objects that is C presuperscript n subscript rfor finding the ways in which we can select two chairs for women and three for men. Then we will permute the men and women among themselves.

The correct answer is: None of these


    DETAILED SOLUTION
    Now, we have been given 8 chairs which are numbered from 1 to 8. Also, it has been given that women choose the chairs from amongst the chairs marked 1 to 4, and then men select from remaining chairs.
    In total there are 2 women and three men who wish to occupy one chair each.

    Now, we know the number of ways of selecting r objects among n is C presuperscript n subscript r. So, we have the ways in which we can choose two chairs among four numbered 1 to 4 is C presuperscript 4 subscript 2 and we can arrange the women then in 2! ways. Also, we have the ways of selecting 3 chairs among the rest 6 chairs is C presuperscript 6 subscript 3 and in them we can permute the men in 3! ways.

    So, in total we have number of possible arrangements as,
    C presuperscript 4 subscript 2 2! cross times C presuperscript 6 subscript 3 3!

    Now, we know that  C presuperscript n subscript r = P presuperscript n subscript r space. space r factorial
    Therefore, we have,

    Total ways = P presuperscript 4 subscript 2 cross times P presuperscript 6 subscript 2

    It is important to note that we have used a fact that C presuperscript n subscript r = P presuperscript n subscript r space. space r factorial. This can be understood as we know that C presuperscript n subscript r = fraction numerator n factorial over denominator left parenthesis n minus r right parenthesis factorial space r factorial end fractionand P presuperscript n subscript r = fraction numerator n factorial over denominator left parenthesis n minus r right parenthesis factorial space end fraction  . So, substituting this we have C presuperscript n subscript r = P presuperscript n subscript r space. space r factorial . 

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