Asked by Jim
how many 3-digit numbers can be formed using only the digits 1 to 7, if the number 2 must be included? (Repetitions are allowed)
The books says 127. Could someone please explain this? Thanks
The books says 127. Could someone please explain this? Thanks
Answers
Answered by
Reiny
Since repetition is allowed, and the 2 must be included, it is easy to forget that the 2 can show up once, twice or even three times
case 1: only one 2
number of ways = 3(6)(6) = 108
(2XX, X2X, XX2, where X is one of the other 6 digits)
case 2: two 2's
number of ways = 3(6) = 18
(22X, 2X2, X22, where X is one of the other 6 digits)
case 3: three 2's , namely 222
number of ways = 1
total = 108+18+1 = 127
case 1: only one 2
number of ways = 3(6)(6) = 108
(2XX, X2X, XX2, where X is one of the other 6 digits)
case 2: two 2's
number of ways = 3(6) = 18
(22X, 2X2, X22, where X is one of the other 6 digits)
case 3: three 2's , namely 222
number of ways = 1
total = 108+18+1 = 127
Answered by
Mgraph
127=7^3-6^3
7^3 all 3-digit numbers, formed using
1,2,3,4,5,6,7
6^3 ... using 1,3,4,5,6,7
7^3 all 3-digit numbers, formed using
1,2,3,4,5,6,7
6^3 ... using 1,3,4,5,6,7
Answered by
Reiny
Mgraph
Very nice, good logic.
Very nice, good logic.
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