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## Competency in Focus: combinatorics

This problem from B.Stat. (Hons.) Admission Test 2005 – Objective Problem 1  is based on arranging integers to get particular results.

## Next understand the problem

1. How many three-digit numbers of distinct digits can be formed by using the digits 1, 2, 3, 4, 5, 9 such that the sum of the digits is at least 12? (A) 61 (B) 66 (C) 60 (D) 11
##### Source of the problem
B.Stat. (Hons.) Admission Test 2005 – Objective problem 1

### Combinatorics

4/10
##### Suggested Book
Challenges and Thrills in Pre College Mathematics Excursion Of Mathematics

Do you really need a hint? Try it first!
Out of $n$ things we can select $r$ thing ins in \left( \begin{array}{c} n \\ r \end{array} \right) ways. Similarly we can select  $3$ integers out of $6$ in \left( \begin{array}{c} 6 \\ 3 \end{array} \right) ways.
We want to count the cases that do not work (That means sum becomes less than 12). Note that if 9 is one of the three digits then the sum of the digits will be 12 or more (least case is 9+1+2 =12).  an similarly you can thing next posibiily.
Note that if we want get the numbers whose sum is equal to or greater than 12 then from the digits 1,2,3,4 and 5 (but not 9), then there is only one posibility and that is  $5+4+3$.
Therefore we select 3 distinct digits from the five digits and delete the one selection of {5, 4, 3}. This gives us the total number of bad cases. $\left( \begin{array}{c} n \\ r \end{array} \right)–1=9$.
Now we have total 20 posibilities out of which 9 are non-working solutions for our question. So total orking solutions is $20-9=11$.  Further we can arrange this 3 letters in $3! =6$ ways.  Hence the total number of required ways =  $11 * 6=66$.

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