How many numbers between 99 and 1000 (both excluding) can be formed such that
a) every digit is either 3 or 7.
b) there is no restriction.
c) no digit is repeated.
d) the digit in hundred's place is 7.
e) the digit 7 does not appear at any place.
f) the digit in units place is 7.
g) atleast one of the digits is 7.
Answers
Step-by-step explanation:
(i) First note that all these numbers have three digits. 7 is in the unit’s place. The middle digit can be any one of the 10 digits from 0 to 9. The digit in hundred’s place can be any one of the 9 digits from 1 to 9. Therefore, by the fundamental principle of counting, there are 10 × 9 = 90 numbers between 99 and 1000 having 7 in the unit’s place. (ii) Total number of 3 digit numbers having atleast one of their digits as 7 = (Total numbers of three digit numbers) – (Total number of 3 digit numbers in which 7 does not appear at all). = (9 × 10 × 10) – (8 × 9 × 9) = 900 – 648 = 252.Read more on Sarthaks.com - https://www.sarthaks.com/116915/i-how-many-numbers-are-there-between-99-and-1000-having-7-in-the-units-place?show=116919#a116919
Consider first that each of these numbers has three digits.
The unit is replaced by 7. Any one of the 10 digits, from 0 to 9, can be used as the middle digit. Any one of the nine digits from 1 to 9 may be used as the digit in the hundredth place.
According to the basic rule of counting, there are 10 * 9 = 90 numbers between 99 and 1000 with 7 in the place of the unit.
The total number of three-digit numbers with at least one seven equals the total number of three-digit numbers minus the total number of three-digit numbers with no seven at all.
= 9 × 10 × 10 – 8 × 9 × 9= 900 – 648 = 252.
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