 Quandaries and Queries hi i am behzad and from iran i am a teacher i saw the question below in the internet but i didn't manage to get the exact answer that was in the site i think the level of the question is secondary(?)   How many six-digit numbers contain at least one 7 in their decimal expansion?   i'll be much more thankful if you intruduce sites with questions and answers about probability   thanks a lot        BEHZAD Hi Behzad, If you are going to form a six digit number from the ten digits 0,1,2,3,4,5,6,7,8,9 then the first digit can't be 0, so you have 9 choices for the first digit. Each of the remaining digits can be any of the 10 digits and hence there are 9 10 10 10 10 10 = 900 000 six digit numbers. Rather than count the number of such six digit numbers that contain at least one 7 I am going to count the number that have no 7's. Hence I want to count the six digit numbers that can be formed from the digits 0,1,2,3,4,5,6,8,9. As before the first digit can't be 0 so there are 8 choices for the first digit. Each of the remaining digits can be any of the 9 digits and hence there are 8 9 9 9 9 9 = 472 392 six digit numbers that contain no 7. The remaining six digit numbers contain at least one 7 and hence the number of six digit numbers that contain at least one 7 is 900 000 - 472 392 = 427 608 Penny Go to Math Central