Help me to verify some Probability answers

[juanitotruan77]

Homework Statement

So i have this 4 problems that i already answered, but i want to know if I am right.
1.- how many 4 letter words can be made with the letters of the word "emulsion" if you can use 2 vocals and 2 consonants.
My answer: 864 4*4*3*3*3!
2._ how many triangles can you make with 8 stones if you can't put more that 3 of them in a straight line
My answer: 8C3 = 56
3._ Find the term that has an x¹² in (-.333x^3+.666y^2)^8
My answer: .86x^12 y^8
4._ If i want to arrange 5 history books, 4 math books and 3 physics books in a shelve, how many ways are there to do it if i want all the books from the same topic in together?
My answer: 5!*4!*3!*3!= 103680

 

[Ray Vickson]

Homework Statement

So i have this 4 problems that i already answered, but i want to know if I am right.
1.- how many 4 letter words can be made with the letters of the word "emulsion" if you can use 2 vocals and 2 consonants.
My answer: 864 = 4*4*3*3*3! <--note "=" added in edit!
2._ how many triangles can you make with 8 stones if you can't put more that 3 of them in a straight line
My answer: 8C3 = 56
3._ Find the term that has an x¹² in (-.333x^3+.666y^2)^8
My answer: .86x^12 y^8
4._ If i want to arrange 5 history books, 4 math books and 3 physics books in a shelve, how many ways are there to do it if i want all the books from the same topic in together?
My answer: 5!*4!*3!*3!= 103680

1) I get your answer 864, but I do not understand how you got it. Assuming you cannot use the same vowel or the same consonant twice, there are C(4,2)*C(4,2) ways to choose the vowels and consonants; then, for each such choices there are 4! permutations of the four selected letters, for a total of C(4,2)^2*4! = 864.
2) I did not do this one.
3) I cheated and used the computer algebra system Maple:
> p:=(2/3*y^2-1/3*x^3)^8; <--input
(2/3*y^2-1/3*x^3)^8 <--input echo
> coeff(p,x,12); <--- find the coefficient of x^12 in p
(1120/6561)*y^8 <---answer
evalf(%); <--- do floating-point evaluation
.1707056851*y^8 (coeffient of x^12)

So, I get 0.17 x^12 y^8, not your 0.86 x^12 y^8.

We can also see this directly: in order to get x^12 we need to use the factor (-1/3)x^3 four times, so we need to use the other factor (2/3) y^2 four times, giving us y^8. To get the numerical coefficient, note that we can choose the four factors containing x in C(8,4) ways, and for each such way we have (2/3)^4*(-1/3)^4, giving coefficient = C(8,4)*(2/3)^4*(-1/3)^4 = 1120/6561. (Strictly speaking, I assumed that .666 really meant 2/3 and .333 really meant 1/3. If those assumptions are false--meaning that .333 and .666 are exact as stated---the coefficient would be C(8,4)*(.666)^4 * (-.333)^4 = .1693448098; but this is still a lot different than your 0.86.

4) Your answer is perfect, but I would have preferred a word or two of explanation: there are 3! ways to arrange the subjects, and there are 5! ways to arrange the history books, etc.

 

[juanitotruan77]

1) I get your answer 864, but I do not understand how you got it. Assuming you cannot use the same vowel or the same consonant twice, there are C(4,2)*C(4,2) ways to choose the vowels and consonants; then, for each such choices there are 4! permutations of the four selected letters, for a total of C(4,2)^2*4! = 864.
2) I did not do this one.
3) I cheated and used the computer algebra system Maple:
> p:=(2/3*y^2-1/3*x^3)^8; <--input
(2/3*y^2-1/3*x^3)^8 <--input echo
> coeff(p,x,12); <--- find the coefficient of x^12 in p
(1120/6561)*y^8 <---answer
evalf(%); <--- do floating-point evaluation
.1707056851*y^8 (coeffient of x^12)

So, I get 0.17 x^12 y^8, not your 0.86 x^12 y^8.

We can also see this directly: in order to get x^12 we need to use the factor (-1/3)x^3 four times, so we need to use the other factor (2/3) y^2 four times, giving us y^8. To get the numerical coefficient, note that we can choose the four factors containing x in C(8,4) ways, and for each such way we have (2/3)^4*(-1/3)^4, giving coefficient = C(8,4)*(2/3)^4*(-1/3)^4 = 1120/6561. (Strictly speaking, I assumed that .666 really meant 2/3 and .333 really meant 1/3. If those assumptions are false--meaning that .333 and .666 are exact as stated---the coefficient would be C(8,4)*(.666)^4 * (-.333)^4 = .1693448098; but this is still a lot different than your 0.86.

4) Your answer is perfect, but I would have preferred a word or two of explanation: there are 3! ways to arrange the subjects, and there are 5! ways to arrange the history books, etc.

Oh i wrote the #3 wroing, It's -.5x^3 not .333x^3.
#4 that's exactly what i did, sorry for the lack of explanation.
#1 is because you can't repeat words, so you have 4*4*3*3 and you have 6 posible orders:
v v c c
v c v c
v c c v
c v v c
c v c v
c c v v
or 6!.

Or you can apply the formula :
spaces!/vowels!consonants!