counting numbers...

• November 14th 2011, 10:31 AM
slapmaxwell1
counting numbers...
ok so here goes....

the digits 3, 6, 7 and 8 will be used without repetition to form different 3 digit numbers. of all such numbers how many are greater then 500?

A. 6
B. 12
C. 18
D. 24
E. 48

i just dont know how to do this problem without writing out all the possible combinations, but on a timed test this wouldn't work because, i would run out of time.

i guessed 12 well 6 was too small, 48 and 24 are too big so it had to be 12 or 18. so i reasoned 4 numbers in 3 digit combinations 4 times 3 = 12?

sat may 2011 section 4 problem 11
• November 14th 2011, 10:34 AM
Plato
Re: counting numbers...
Quote:

Originally Posted by slapmaxwell1
the digits 3, 6, 7 and 8 will be used without repetition to form different 3 digit numbers. of all such numbers how many are greater then 500?

$(3)(3)(2)=~?$
• November 14th 2011, 10:36 AM
Deveno
Re: counting numbers...
the numbers must begin with 6,7,or 8.

that's 3 possibilities.

after choosing the first digit (that is, for each of these 3 possibilities), how many more digits do we have to choose, and how many ways can we choose them?
• November 14th 2011, 12:15 PM
slapmaxwell1
Re: counting numbers...
Quote:

Originally Posted by Plato
$(3)(3)(2)=~?$

ok but where does the 2 come from...
• November 14th 2011, 12:31 PM
Plato
Re: counting numbers...
Quote:

Originally Posted by slapmaxwell1
ok but where does the 2 come from...

We want to have two conditions fulfilled: greater than 500 all digits distinct.
First number can be chosen in three ways. (greater than 500)
Second number can be chosen in three ways. (all digits distinct)
Third number can be chosen in two ways. (all digits distinct)
• November 14th 2011, 04:58 PM
Soroban
Re: counting numbers...
Hello, slapmaxwell1!

You can think your way through this one . . .

Quote:

The digits 3, 6, 7 and 8 are used without repetition to form 3-digit numbers.
Of all such numbers, how many are greater then 500?

. . (A) 6 . . (B) 12 . . (C) 18 . . (D) 24 . . (E) 48

We are forming 3-digit numbers:. $\begin{array}{c}\sqcap\;\sqcap\;\sqcap \\ [-3mm]\sqcup\;\sqcup\;\sqcup \end{array}$

The first digit must be 6, 7, or 8 . . . 3 choices.

The second digit must one of the other three digit . . . 3 choices.

The third digit must be one of the other two digits . . . 2 choices.

Therefore, there are:. $3\cdot3\cdot2 \,=\,18$ possible 3-digit numbers.