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7.38i don't know how to do these type of questions.
Shot at 2007-08-08
2357 arranged so that the number is < 4000. That means the 2 or 3 is in front. So the first digit is one of two (2 or 3). The second digit can be any of the 3 remaining ones, the third digit can be any of the 2 remaining ones, and the last digit is the remaining one. So you have 2*3*2*1 = 12 ways.
7.39
Even 4-digit numbers using 1239, no repeats. So the 2 has to be the fourth digit because the rest are odd. That means any digit can be in any other spot. Any of the 3 can be in the first spot, any of the remaining 2 can be in the second spot, and you have 1 left for the third spot. That's 3*2*1 = 6 ways.
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Ok, I think I know how to do this, correct me if I am wrong Streetwolf, cuz I know u know this one...
First, the number must be under 4000, therefore it must start with 2 or 3. Assuming you must use all of the numbers in the problem to create a value, then you have 2XXX and 3XXX.
Create a table if you like or remember that this is a permutation.
3 nPr 3=6 (also can do 3!, which is 3x2x1)
which means you can have six different values for either 2 or 3 for a total of 12 different values.
Assuming that you DON'T HAVE TO USE ALL 4 NUMBERS to create a value, you do the same AND...
any 3 numbers can be put together to be less than 4000, so do another permutation...
4 nPr 3=24 (say it " 4 choose 3", done as 4x3x2, e.g.- 237, 573, etc...)
and
4 nPr 2=12 (4X3, e.g.- 57, 72, etc...)
then you have each number itself...
2,3,5,7 for a total of 4 more.
add them together, 12+12+24+4=52
First, the number must be under 4000, therefore it must start with 2 or 3. Assuming you must use all of the numbers in the problem to create a value, then you have 2XXX and 3XXX.
Create a table if you like or remember that this is a permutation.
3 nPr 3=6 (also can do 3!, which is 3x2x1)
which means you can have six different values for either 2 or 3 for a total of 12 different values.
Assuming that you DON'T HAVE TO USE ALL 4 NUMBERS to create a value, you do the same AND...
any 3 numbers can be put together to be less than 4000, so do another permutation...
4 nPr 3=24 (say it " 4 choose 3", done as 4x3x2, e.g.- 237, 573, etc...)
and
4 nPr 2=12 (4X3, e.g.- 57, 72, etc...)
then you have each number itself...
2,3,5,7 for a total of 4 more.
add them together, 12+12+24+4=52
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I think for the first one, you can just think of it like this... for the number to be less than 4000, the first digit has to be either 2 or 3, so you start with two possible numbers there. The next three digits can be any of the others, so you end up with 2*3*2*1 = 12.
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