renriqu1

They hardly let you bring anything. I know no writing instruments, but ear plugs are allowed, no earplug case though.

Thanks a lot Danielman, that's what I was looking for. Could you or calm_J elaborate on the part of 4! for part (a). I still don't understand why it is 4! for the possible combinations, but I understand why and how much is subtracted.

The thing I'm confused is about is that last time I did 3x3x3 - 1 (for the one that was not possible). So in this case why isn't it 4x4x4x4 - the none possible ones?

I posed questions not statements. Thanks for the enlightenment.

The thing is, there is no distinction between the 3 different red balls. If there was a certain marking, like 1, 2, and 3, then it would be a different problem?

If Q1 asked how many different ways are there to make the 3 digit number out of those numbers, would it not be the same as asking how many different numbers are there?

Thanks calm_J The way you're doing the first question it first appears that you're not letting any of the letters repeat. 4! would be like you used T, now you're only left with 3 letters, E, N, and S, then 2, and then just one. You got the right answer, but I still don't understand your method. The answer to Q2 still stands as 163. This may be a typo, but I don't know.

Hey guys, here's a really fast way to look at it. Use the quadratic equation: For the example here: x^2 + ax + b, x = 3 + sqrt(2). x is therefore [-a +/- sqrt(a^2 - 4*1*b)]/(2*1) Take note of the brackets and parantheses Then do it by parts...first solve for a. -a/2 = 3 -----> a = -6 Now plug this into the second part.. sqrt((-6)^2 - 4b)]/2 = sqrt(2) sqrt(36 - 4b)] = 2 sqrt(2) = (sqrt(4))*(sqrt(2)) = ((sqrt(8)) Now drop the square roots... 36 -4b = 8 4b = 28 -----> b = 7

Could help me with this one? I don't know how to get this one at all. Thanks in advance. A 4 letter word can be made from the letters of the word TENNESSEE. Find the number of: (a) combinations (b) permuatations Answers: (a) 17 (b) 163

Q2 has 3 parts. These questions came from a probability and statistics book so they don't have choices like the GREs, but are similar to questions previously posted on this forum. For question 2: a) no restrictions imposed. There are 6 people in a circle. 1 person sits down, and then there are 5! ways to sit the rest of the people. The answer is 5! = 120 b) I found this by subtracting the number of ways 2 women can sit together from the total number of ways to sit in the circle. So the two women sitting together can be seen as a pair, so consider there are now only 5 people in the circle. There are 4! ways to arrange them, but there are also 2! ways to arrange the two women. The number of ways for two women to sit together is 4! x 2! = 48. Now subtarct this from the total ways of arranging people (120) and you will get 72. c) Imagine a circle where it has to be man, woman, man, woman, etc. Each woman has 3 spots, each man has 3 spots. Say one woman sits down, then there are 2! ways for the other women to sit down and 3! ways for the men to sit. The answer is 3! x 2! = 12.

I found the answer using this method: For each digit there are 3 choices, however, the combination of 333 is not possible. This leads to (3x3x3)-1 = 26.

Here are some permutation problems: 1) How many different three-digit numbers can be made with 3 fours, 4 twos, and 2 threes? 2) In how many ways can 3 men and 3 women be seated at a round table if: a) no restriction is imposed b) 2 particular women must not sit together c) each woman is between 2 men? 1) 26 2) a) 120 b) 72 c) 12