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Arrangements

Expert replies
by yellowho » Fri Jan 21, 2011 12:48 am
How many integers from 100,000 to 1,000,000 are such that each integer contains no repeated
digits, and the integer’s digits are arranged in ascending order from least to greatest?

How do you restrict the order?
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Source: — Problem Solving |

by Rahul@gurome » Fri Jan 21, 2011 1:13 am
yellowho wrote:How many integers from 100,000 to 1,000,000 are such that each integer contains no repeated
digits, and the integer’s digits are arranged in ascending order from least to greatest?
In other words, the question is asking how many 6 digit integers in which all the integers are different and they are arranged in ascending order from least to greatest.

Now note that only one arrangement is possible for any combination of 6 different integers such that they are arranged in ascending order from least to greatest. Thus the problem boils down to selection of 6 different integers out of 9 integers. We are not taking 10 integers as zero cannot be considered as a digit because if we consider zero as a digit, it must be placed in the beginning which will result a 5 digit number.

Hence, number of such integers = Number of possible selections of 6 different digits out of 9 = 9C6 = 84
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by GMATGuruNY » Fri Jan 21, 2011 3:25 am
yellowho wrote:How many integers from 100,000 to 1,000,000 are such that each integer contains no repeated
digits, and the integer’s digits are arranged in ascending order from least to greatest?

How do you restrict the order?
Even though the problem uses the word arranged, this is not an arrangement question but a combination question. Given any combination of 6 different digits, there is only 1 way to arrange them from smallest to greatest. Thus, any combination of 6 different digits will yield 1 possible arrangement.

So all we need to do is count the number of combinations of 6 that be formed from the 9 digits: 9C6 = 84.
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