Noxy416 wrote:
Of the three-digit positive integers whose three digits are all different and nonzero, how many are odd integers greater than 700?
(A) 84
(B) 91
(C) 100
(D) 105
(E) 243
What am i doing wrong in this question?
Number of options for the hundreds digit =3
Number of options for the Units digit =4
Number of options for the Units digit = 8
Total number of options- 3 x 8 x 4 = 96
Please help
Please follow posting guidelines (link in my signatures). Do make sure to mention the correct title and put your analyses either in the next post or under spoilers.As for the question, you need to find the number of combinations possible such that:
1. It is a 3 digit number > 700 (or between 701-999, inclusive).
2. All digits are different and NON ZERO.
3. The numbers must be ODD --> the last digit can be 1 of 1,3,5,7,9
Based on this, the numbers can be of the following 3 types:
Type 1: 7AB
Type 2: 8EF
Type 3: 9CD
For type 1 and type 3, be very careful that the digits must be different. So if it is 7AB, then B can NOT be 7. Similarly for type 3, 9CD, D can NOT be 9. There is no such restriction when you find numbers of type 2 (8EF).
Number of combinations for type 1 : 1*7*4 = 28
Number of combinations for type 2: 1*7*5 = 35
Number of combinations for type 3 : 1*7*4 = 28
Thus, total numbers possible = 28+35+28 = 91.
B is thus the correct answer.
Hope this helps.
We have total 9 digits. One is used in hundreds place, then we must be left with 8 digits for tens place. Please help me understand where am I going wrong