Permutations And Combinations

Question
CBSEENMA11014121

There are three prizes to be distributed among 6 students. In how many ways can this be done when

(a) no boy gets more than one prize.

(b) there is no restriction as to the number of prizes that a boy may get.

(c) no boy sets all prizes.

Solution

(a) n = 6, r = 3
Number of permutations = straight P presuperscript straight n subscript straight r space equals space straight P presuperscript 6 subscript 3 space equals space fraction numerator 6 cross times 5 cross times 4 cross times 3 factorial over denominator 3 factorial end fraction space equals space 120.
Hence, the number of ways, in which the three prizes can be awarded = 120.
(b) n = 6,  r = 3
Number of permutations = straight n to the power of straight r space equals space 6 cubed space equals space 216.
Hence, the number of ways in which 3 prizes can be awarded = 216.
(c) Number of ways in which one boy gets all prizes = number of boys = n = 6.
                      equals straight n to the power of straight r space minus straight n space equals space 6 cubed space minus space 6 space equals space 216 space minus space 6 space equals space 210.

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