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Exam pack 2 permutation How to solve?

Expert replies
Source: — Problem Solving |

by [email protected] » Tue May 24, 2016 10:52 pm
Hi Neilsheth2,

You'll likely find it easiest to handle this calculation in 'pieces'

This question lays out the following restrictions:
1) 3-digit numbers greater than 700
2) All digits are NON-0 and DIFFERENT
3) The number must be ODD.

Let's start with the 700s...

1st digit is 7 = 1 option
3rd digit must be ODD, but NOT 7 = 4 options
2nd digit must differ from the other two digits and not be 0 = 7 options
(1)(4)(7) = 28 numbers in the 700s

Now, the 800s...

1st digit is 8 = 1 option
3rd digit must be ODD = 5 options
2nd digit must differ from the other two digits and not be 0 = 7 options
(1)(5)(7) = 35 numbers in the 800s

Finally, the 900s; the math here works the same as the 700s...

1st digit is 9 = 1 option
3rd digit must be ODD, but NOT 9 = 4 options
2nd digit must differ from the other two digits and not be 0 = 7 options
(1)(4)(7) = 28 numbers in the 900s

28+35+28 = 91

Final Answer: B

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by Neilsheth2 » Tue May 24, 2016 11:41 pm
Thanks got it !
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by GMATGuruNY » Wed May 25, 2016 4:30 am
Neilsheth2 wrote:Of three digit positive integers whose three digits are all different and non zero, how many are odd integers greater than 700

1)84
2)91
3)100
4)105
5)243
Case 1: Hundreds digit is 7 or 9
Number of options for the hundreds digit = 2. (7 or 9)
Number of options for the units digit = 4. (Any odd digit but the one in the hundreds place)
Number of options for the tens digit = 7. (Any digit 1-9 but the two already used)
To combine these options, we multiply:
2*4*7 = 56.

Case 2: Hundreds digit is 8
Number of options for the hundreds digit = 1. (Must be 8)
Number of options for the units digit = 5. (Any of the 5 odd digits)
Number of options for the tens digit = 7. (Any digit 1-9 but the two already used)
To combine these options, we multiply:
1*5*7 = 35.

Total options = 56+35 = 91.

The correct answer is B.
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