# Permutation and Combination

1. Jan 12, 2013

### e.pramudita

1. The problem statement, all variables and given/known data
If three different number are taken from the set {0, 1, 3, 5, 7} to be used as the coefficient of a standard quadratic equation, how many such quadratic equations can be formed? How many of these have real roots?

2. Relevant equations

3. The attempt at a solution
Part a: "How many such quadratic equations can be formed?"

Quadratic equation is defined by ax^2 +bx +c =0 , a is not 0
First box (x^2) = 4 (consist of everything but zero)
Second box (x^1) = 4 (consist of everything but the number on the first box)
Third box (x^0) = 3 (consist of everything but the number of the first box and the second box)
So 4*4*3=48.

Part b: "How many of these have real roots?
Imaginary roots have D<0
So lets find how many equation that have imaginary roots and then subtract Part(a) answer with that number.

Case 1: x^2 -> non zero -> (4)
x^1 -> {0,1,3} -> (3)
x^0 -> {non zero except the first and second box} -> (3)
So 4*3*3=36

Case 2: x^2 -> non zero -> (4)
x^1 -> {5} -> (1)
x^0 -> {3,7} ->(2)
So 4*1*2=8

I can think of 2 other cases but let stop for a second here.
From here on we can conclude that there is at least 36+8=44 combination. Thus there are 48-44=4 quadratic equation that have real roots.

But I can think of at least 12 real roots.
Set coefficient x^0 = 0 and other coefficient with the rest.
We can get.
4*3*1=12

I am confused!

2. Jan 12, 2013

### thchian

the standard quadratic equation is f(x) = a(x - h)2 + k

3. Jan 12, 2013

### e.pramudita

Can you give more explanation on how to solve it?

4. Jan 12, 2013

### CAF123

Taking the quadratic to be of the form $ax^2 + bx + c$, $a \neq 0$, using a slightly different method to yours, I also get 48.

5. Jan 12, 2013

### thchian

IF the number that can be used is only 0, 1, 3, 5, 7, without any restriction, the numbers of equations that can be formed at most are only 5P3=60, right?
So, do u think it means 10, 37, 137,... those numbers can be be used?

6. Jan 12, 2013

### CAF123

There is restriction in that a cannot be 0. I think it means three numbers from the given set( I.e you cannot combine them to form larger digit numbers)

7. Jan 12, 2013

### e.pramudita

There is a restriction

8. Jan 12, 2013

### e.pramudita

9. Jan 12, 2013

### thchian

i mean if no restriction answer = 60, then when there is a restriction answer will be < 60...

10. Jan 12, 2013

### CAF123

For real root quadratics, you have $b^2 - 4ac \geq 0\,\Rightarrow\,b^2 \geq 4ac$ (valid because $a,c$ both positive). How many combinations of a,b,c satisfy this?

Last edited: Jan 12, 2013
11. Jan 12, 2013

### haruspex

For the second part, consider the cases in this order:
c = 0
c > 0, b = 0 or 1
c > 0, b = 3
c > 0, b > 3

12. Jan 13, 2013

### CAF123

For part a), note that the answer cannot possibly be 88. As thchian pointed out, there would be, without any restrictions, 5P3 = 60 possibilities so with some restrictions this number must be less than 60. To see that it is necessarily 48 again, count the number of combinations which result in a being 0. This gives 4P2 = 12. So subtracting gives 60 -12 = 48.