Permutations And Combinations

Question
CBSEENMA11014038

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?

Solution


(a) Number of digits available = 6

Number of places [(x), (y) and (z)] for them = 3
Repetition is allowed and the 3-digit numbers formed are odd
Number of ways in which box (x) can be filled = 3 (by 1, 3 or 5 as the numbers formed are to be odd)
rightwards double arrow               m = 3
Number of ways of filling box (y) = 6                           (∴ Repetition is allowed)
rightwards double arrow               n = 6
Number of ways of filling box (z) = 6                           (∵ Repetition is allowed) 
rightwards double arrow              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 )  
rightwards double arrow                                   m = 3
Number of ways of filling box (y) = 5                                (∵ Repetition is not allowed)
rightwards double arrow                              n = 5
Number of ways of filling box (z) = 4                                 (∵ Repetition is not allowed)
rightwards double arrow                             p = 4
∴     Total number of 3-digit odd numbers formed
                                  = m x n x p = 3 x 5 x 4 = 60.   

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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.