Question

Please answer entire question

Problem 2. (a) How many three-digit numbers can be formed from the digits 0, 1, 2, 3, 4, 5, and 6 if each digit can be used only once? (Note: 062 is NOT a legit three-digit number) (b) How many of these are odd numbers? (c) How many are greater than 330?

Probabilistic Models in Industrial Engineering

Answer 1

a) number of ways =N(for first digit we have 6 choices except that of 0, for 2nd remaining 6 choices (including 0) and for 3rd remaining 5 choices)

=6*6*5 =180

b)

number of odd digits =N(last digit is odd for which we have 3 choices, for 1st digit 5 choices (except last and 0) , for 2nd digit remaining 5 choices) =3*5*5 =75

c)

number of ways =N(1st is 3 , 2nd is 4 or more for which 3 choices and for third 5 choices)+N(1st is 4 or more for which 3 choices, for 2nd remaining 6 and for third remaining 5 choices)

=1*3*5+3*6*5 =105