Indian Statistical Institute B.Math & B.Stat
How many $three-digits$ numbers of distinct digits can be formed by using the digits $1,2,3,4,5,9$ such that the sum of digits is at least 12? $$$$
First note that since the digits must be distinct there must be no repetition and further note that the maximum sum of such three digits is $9+5+4=18.$ $$$$
We will find all such possible by considering their sum of the digits. $$$$
Case I: Sum of the digits is 12. In this case the selection of digits can be \(\{1,2,9\},\{3,4,5\}\) $$$$
So, a total of $3!+3!=12$ $three-digits$ numbers possible. $$$$
Case II: Sum of the digits is 13. In this case the selection of digits can be \(\{1,3,9\}\) $$$$
So, a total of $3!=6$ $three-digits$ numbers possible. $$$$
Case III: Sum of the digits is 14. In this case the selection of digits can be \(\{1,4,9\},\{2,3,9\}\) $$$$
So, a total of $3!+3!=12$ $three-digits$ numbers possible. $$$$
case IV: Sum of the digits is 15. In this case the selection of digits can be \(\{1,5,9\},\{2,4,9\}\) $$$$
So, a total of $3!+3!=12$ $three-digits$ numbers possible. $$$$
Case V: Sum of the digits is 16. In this case the selection of digits can be \(\{2,5,9\},\{2,5,9\}\) $$$$
So, a total of $3!+3!=12$ $three-digits$ numbers possible. $$$$
Case VI: Sum of the digits is 17. In this case the selection of digits can be \(\{3,5,9\}\) $$$$
So, a total of $3!=6$ $three-digits$ numbers possible. $$$$
Case VII: Sum of the digits is 18. In this case the selection of digits can be \(\{4,5,9\}\) $$$$
So, a total of $3!=6$ $three-digits$ numbers possible. $$$$
Adding all the cases we have $12+6+12+12+12+6+6=66$, $three-digits$ numbers of distinct digits can be formed.
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