How many 3-digit odd numbers can be formed from the digits 1,2,3,4,5,6 if:
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.
Solution not provided.Ans. (i) 720 (ii) 1440 (iii) 576 166 Views
Discussion :: Permutation and Combination - General Questions (Q.No.2)
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