How many ways 4 prizes given away to 3 students all are equally eligible?

How many 3-digit odd numbers can be formed from the digits 1,2,3,4,5,6 if:
(a) the digits can be repeated (b) the digits cannot be repeated?

How many ways 4 prizes given away to 3 students all are equally eligible?
(a) Number of digits available = 6

Number of places [(x), (y) and (z)] for them = 3Repetition is allowed and the 3-digit numbers formed are oddNumber of ways in which box (x) can be filled = 3 (by 1, 3 or 5 as the numbers formed are to be odd)

               m = 3

Number of ways of filling box (y) = 6                           (∴ Repetition is allowed)

               n = 6

Number of ways of filling box (z) = 6                           (∵ Repetition is allowed) 

              p = 6

∴  Total number of 3-digit odd numbers formed                             = m x n x p = 3 x 6 x 6 = 108(b) Number of ways of filling box (x) = 3                     (only odd numbers are to be in this box )  

                                   m = 3

Number of ways of filling box (y) = 5                                (∵ Repetition is not allowed)

                              n = 5

Number of ways of filling box (z) = 4                                 (∵ Repetition is not allowed)

                             p = 4

∴     Total number of 3-digit odd numbers formed

                                  = m x n x p = 3 x 5 x 4 = 60.   

(i) Since no student gets more than one prize; the first prize can be given to any one of the five students.The second prize can be given to anyone of the remaining 4 students.Similarly, the third prize can be given to any one of the remaining 3 students.The last prize can be given to any one of the remaining 2 students.

    ∴ Required number of ways =`5xx4xx3xx2=5!`

(ii) Since a student may get any number of prizes, the first prize can be given to any of the five students. Similarly, the rest of the three prizes can be given to the each of the remaining 4 students.
∴ Required number of ways =`5xx5xx5xx5=625`

(iii) None of the students gets all the prizes.
  ∴ Required number of ways = {Total ways of distributing the prizes in a condition wherein a student may get any number of prizes - Total ways in a condition in which a student receives all the prizes} =`625-5=620`

sir iss question m bola gya h.. that if each boy is eligible for ALL the prize... then the total ways must be 4*4*4 because 1st boy can get all the prize similarly 2nd boy nd 3rd boy too

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