## What are we learning ?

**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.

## First look at the knowledge graph:-

## 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

##### Key Competency

##### Difficulty Level

4/10

##### Suggested Book

Challenges and Thrills in Pre College Mathematics Excursion Of Mathematics

## Start with hints

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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