Permutations And Combinations

Question
CBSEENMA11014197

In how many ways can a team of 3 boys and 3 girls be selected out of 5 boys and 4 girls? 

Solution

Number of boys = 5
Number of boys to be selected = 3
rightwards double arrow  Number of combinations (selections) = straight C presuperscript straight n subscript straight r space equals space straight C presuperscript 5 subscript 3 space equals space fraction numerator 5 factorial over denominator 3 factorial space 2 factorial space end fraction space equals space 10      ...(i)
                       Number of girls = 4
        Number of girls to be selected  = 3
rightwards double arrow                   Number of selections = straight C presuperscript straight n subscript straight r space equals space straight C presuperscript 4 subscript 3 space equals space fraction numerator 4 factorial over denominator 3 factorial space 1 factorial end fraction equals 4              ...(ii)
∴ From (i) and (ii), by fundamental principle of counting, the number of selections
                                             = 10 x 4 = 40
Hence, the number of ways in which 3 boys and 3 girls out of 5 boys and 4 girls can be selected = 40
        
 

Sponsor Area

Some More Questions From Permutations and Combinations Chapter

Determine K, so that K + 2, 4K – 6 and 3K – 2 are three consecutive terms of an A.P.