How many integers are there from 499 to 999 b How many odd i
Solution
a) No. of integers from 499 to 999 = 999 - 499 + 1 = 501
b) Odd integers from 100 to 999 are 101, 103,.....,999
a = 101, d = 2 and an = 999
So by Arithmetic progression, an = a + (n - 1)d = 101 + 2(n - 1)
=> 999 - 101 = 2(n - 1)
=> n = 450
c) At hundredth place we have 9 choices of placing a no. because 0 can\'t come. On tenth place we have 9 choices again as we can\'t place the number that we have placed at 100th place. Similarly we have 8 choices for unit place.
So no. of distinct digit numbers from 100 to 999 = 9*9*8 = 648
d) At unit place we have 5 choices (1,3,5,7,9). Now at hundredth place we have 8 choices of placing a no. because 0 can\'t come and one of the remaining is occupied at unit place. On tenth place we have 8 choices again as we can\'t place the number that we have placed at 100th place and unit place.
So no. = 8*8*5 = 320
e) 0 at the will not be considered as 3 digit integer.
At hundredth place we have 9 choices of placing a no. because 0 can\'t come. On tenth place we have 9 choices again as we can\'t place the number that we have placed at 100th place. Similarly we have 8 choices for unit place.
So probability = 9/10 * 9/10 * 8/10 = .648
f) 0 at the will not be considered as 3 digit integer.
At unit place we have 5 choices (1,3,5,7,9). Now at hundredth place we have 8 choices of placing a no. because 0 can\'t come and one of the remaining is occupied at unit place. On tenth place we have 8 choices again as we can\'t place the number that we have placed at 100th place and unit place.
So probability = 8/10 * 8/10 * 5/10 = .320

