Combinatorics :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.