Maths question HELP PLEASE!! CARROTSTICKS??? ARE YOU THERE??? (1 Viewer)

[jackerino]

Someone please help here,

"Let S=(a,b,c,d,e,f,g,h,i,j) be the set consisting of the first ten letters of the alphabet."
i)How many subsets of S contain at least three letters?
ii) How many subsets with at least three letters do not contain any vowels?
iii) How many subsets with at least three letters do not contain any vowels but do contain the letter b?


I would greatly appreciate help from one of you maths pro's out there. PLease.

 

[Carrotsticks]

i) Note: I am including the empty set here.

10C3 + 10C4 + ... + 10C10, which is 10C0 + 10C1 + 10C2 + ... + 10C10 - (10C0 + 10C1 + 10C2).

However, do you remember how 10C0 + 10C1 + ... + 10C10 = 2^10?

So therefore 10C3 + 10C4 + ... + 10C10 = 2^10 - (10C0 + 10C1 + 10C2) = 968.

ii) Same thing as (i) except the total is 7 instead of 10, since we have 3 vowels. So we have:

7C3 + 7C4 + 7C5 + 7C6 + 7C7 = 2^7 - (7C0 + 7C1 + 7C2) = 99

iii) Well since it MUST contain the letter B, combinations with 3 letters become combinations for 2 letters (since B must be in it). Combinations with 4 letters become combinations with 3 letters and so on, until combinations of 6 letters (read below for explanation) become combinations of 5 letters.

So now we have a total of 6 letters to pick from. This is because 3 are vowels and we already assume B is in it.

So the total is:

6C2 + 6C3 + 6C4 + 6C5 = 56.

 

[jackerino]

i) Note: I am including the empty set here.

10C3 + 10C4 + ... + 10C10, which is 10C0 + 10C1 + 10C2 + ... + 10C10 - (10C0 + 10C1 + 10C2).

However, do you remember how 10C0 + 10C1 + ... + 10C10 = 2^10?

So therefore 10C3 + 10C4 + ... + 10C10 = 2^10 - (10C0 + 10C1 + 10C2) = 968.

ii) Same thing as (i) except the total is 7 instead of 10, since we have 3 vowels. So we have:

7C3 + 7C4 + 7C5 + 7C6 + 7C7 = 2^7 - (7C0 + 7C1 + 7C2) = 99

iii) Well since it MUST contain the letter B, combinations with 3 letters become combinations for 2 letters (since B must be in it). Combinations with 4 letters become combinations with 3 letters and so on, until combinations of 6 letters (read below for explanation) become combinations of 5 letters.

So now we have a total of 6 letters to pick from. This is because 3 are vowels and we already assume B is in it.

So the total is:

6C2 + 6C3 + 6C4 + 6C5 = 56.

Ahhh thanks so much man! I appreciate it so much

 

[jackerino]

Then there's one which I posted before and noone helped me, am unsure if I approach it correctly,

"if tan(a)/2=tan(b)/3=tan(c)/4=k, show k=sqrt(6)/4"

I did bunch of algebraic manipulation, is that right way to go>

 

[Carrotsticks]

Are you sure you're not missing any pieces of information ie: a,b,c are angles of a triangle?

 

[jackerino]

yea a, b and c are angles of triangle.

 

[jackerino]

Oh and a random question, when R splits up interval PQ into the ratio k:l, does k/l have anything to do with the equation of PQ?

 

[Carrotsticks]

tanA=2k

tanB=3k

tanC=4k

But we know that if A,B,C are angles of a triangle, then tanAtanBtanC = tanA + tanB + tanC.

So 24k^3 = 9k

Now solve for k and note that k can't be 0 or negative.

This is because if A,B,C are angles of a triangle, then suppose we have an obtuse angle. Therefore the other two angles must be acute.

So suppose A is obtuse, then tanA = 2k < 0 (since tan is negative in 2nd quadrant).

Therefore, k is negative.

But tanB = 3k > 0 since B is acute, and same for tanC. This is a contradiction.

Therefore we can also conclude that A,B,C are all acute angles.

 

[Carrotsticks]

Oh and a random question, when R splits up interval PQ into the ratio k:l, does k/l have anything to do with the equation of PQ?

Nothing to do with the equation, since k/l just depends on where R is.

 

[jackerino]

Oh cos in a question I found,

"Point R(x,y) divides P(x1,y1) and Q(x2,y2) into ratio k:l. Show k/l=(y-y1)/(y2-y)."

So I got that bit, then it says,

"Hence show equation of the line PQ is (x-x1)(y-y2)=(y-y1)(x-x2)"

I ignored hence cos I thought it was unnecessary if you already have PQ, but I ended up with a different final equation, so I think you have to somehow use k/l in calculation.

 

[Carrotsticks]

Oh cos in a question I found,

"Point R(x,y) divides P(x1,y1) and Q(x2,y2) into ratio k:l. Show k/l=(y-y1)/(y2-y)."

So I got that bit, then it says,

"Hence show equation of the line PQ is (x-x1)(y-y2)=(y-y1)(x-x2)"

I ignored hence cos I thought it was unnecessary if you already have PQ, but I ended up with a different final equation, so I think you have to somehow use k/l in calculation.

That's a bit different, it's just providing an alternative way of finding the equation of PQ.

 

[jackerino]

That's a bit different, it's just providing an alternative way of finding the equation of PQ.

But, if they specified in the question show THIS is (One of the equations) of PQ, does that mean we have to specifically derive that one? Or is it okay if I have manually done one? anyway how does this "Alternative" way work?

 

[Carrotsticks]

k/l can be found using the Y values or the X values. So similarly for the X values, k/l = (x-x_1)/(x-x_2)

Then equate and move the terms to other side etc.

 

[Kostie100]

What level of maths is this?

 

[Carrotsticks]

3U maths, but a bit harder.