Mathematics: Post your doubts here!

[qwertypoiu]

View attachment 59758
MAJ JUNE 2008
help in I

6 pop, 3 jazz, 2 classical.
So we group jazz together:

PPPPPP[JJJ]CC

The pops and classics can move around but jazz has to stick together (but can move as a group).
So there are 9 elements all together (including Jazz block). (these can be arranged 9! ways)
Note that each of the J's are different from one another. (same goes for P's and C's)
So within the [JJJ] block, the three CD's can move around to form different arrangements. These can permute 3! ways.
So answer = 9! * 3! = 2,177,280

 

[Copy Cat]

If ABC is really a straight line, the points A, B, and C are said to be collinear.

If ABC is really a straight line, the vector AB and the vector BC must be in the same direction. (parallel). So:

AB = OB - OA = [2 3 1]
BC = OC - OB = [1 , p-5 , q+2]

If two vectors are parallel, a common factor can set them equal to one another:

AB = kBC
[2 3 1] = k*[1, p-5, q+2]

It is clear than k=2 here. (By equating the i component of the vector equation)
So now we can equate the other components (j and k):
3 = 2(p - 5)
p = 6.5
1 = 2(q + 2)
q = -1.5

Thanks....

 

[darks]

The general coordinates of any point lying on the two lines may be expressed as:
View attachment 59766
If these two lines are to join, these coordinates need to become equal to each other at some value of λ and μ.
So:

1. 3 - λ = 4 + aμ
2. -2 + 2λ = 4 + bμ
3. 1+λ = 2 - μ

You'll notice the last equation doesn't contain a or b. It can be used to eliminate either λ or μ from the first two equation:
1+ λ = 2 - μ
λ= 1 - μ

So equation 1 becomes (by substituting λ = 1 - μ into it):
3 - 1 + μ = 4 + aμ
μ = 2 + aμ
(1-a)μ = 2
μ = 2/(1-a)

The reason we made μ the subject here will become clear later.

Equation 2 becomes (again after substituting λ = 1 - μ into it):
-2 + 2(1 - μ) = 4 + bμ
-2 + 2 - 2μ = 4 + bμ
(-b-2)μ = 4

Now we can substitute μ from above into here:

(-b-2)*( 2/(1-a) ) = 4
-2b - 4 = 4 - 4a
4a - 2b = 8
2a - b = 4

Part (ii):

If two lines are perpendicular, the dot product of their direction vectors = 0
So:
[-1 2 1] ⋅ [a b -1] = 0
-a + 2b -1 = 0
-a + 2b = 1

Solve this simultaneously with 2a- b = 4 and you'll get:
a = 3, b = 2

Now for the location of their intersection, you just have to substitute these values and do it the usual way:

View attachment 59767
Solve for λ and μ from any two of these three components (check for consistency by substituting into the third component), then obtain the coordinate by substituting either λ or μ into either of the line equations.
You'll get λ = 2, and μ = -1

So the coordinates will be (1,2,3)

Sorry forgot your last question:
For these types of question just apply log to both sides so that they become like linear equations:

5^(2x-1) = 2(3^x)

ln(5^(2x-1)) = ln(2(3^x))
(2x-1)*ln5 = ln2 + x*ln3

(2ln5)x - ln5 = ln2 + (ln3)x
(2ln5 - ln3)x = ln2 + ln5

Do the rest using calculator:
x = 1.085990045

Thank You so much!

 

[Saad the Paki]

Part (iv) and (v) please. The answers are 168 for (iv) and 476 for (v). I didn't quite understand what the question is asking. :/

 

[Rizwan Javed]

View attachment 59783 Part (iv) and (v) please. The answers are 168 for (iv) and 476 for (v). I didn't quite understand what the question is asking. :/

In this question, you're asked to SELECT 3 cards from the 9 given cards. Then you're asked to arrange those 3 cards you've selected in a line.
Part (iii) is simple. You can do that by selecting 3 from 9 (9C3) and then arranging these 3. This will done in 9C3 * 3! OR 9P3 ways.

(iv) In this part, a pink card MUST be selected. So you're to make selections for other two cards only. Select the other two cards from 8 remaining: 8C2. Now arrange the 3 cards (the two you selected from the 8 cards + that must pink card) : 8C2 * 3! = 168

(v) Now you're to find the no. of ways in which a pink card and a green card are not together. You can do this by first finding the no. of arrangements which have pink and green cards together, and then subtracting this from the total possible arrangements with no restrictions.
So first find the arrangements which have pink (P) and green (G) together:
[P,G] O <--- treat Pink and green as a group. P, G can be arranged in 2 ways within the group.

Now make a selection for the other card, O. This will be 7C1. Arrange these 3 cards: 2 * 7C1 * 2 = 28

Subtract 28 from the answer in (iii)

9P3 - 28 = 476 arrangements.

 

[Saad the Paki]

In this question, you're asked to SELECT 3 cards from the 9 given cards. Then you're asked to arrange those 3 cards you've selected in a line.
Part (iii) is simple. You can do that by selecting 3 from 9 (9C3) and then arranging these 3. This will done in 9C3 * 3! OR 9P3 ways.

(iv) In this part, a pink card MUST be selected. So you're to make selections for other two cards only. Select the other two cards from 8 remaining: 8C2. Now arrange the 3 cards (the two you selected from the 8 cards + that must pink card) : 8C2 * 3! = 168

(v) Now you're to find the no. of ways in which a pink card and a green card are not together. You can do this by first finding the no. of arrangements which have pink and green cards together, and then subtracting this from the total possible arrangements with no restrictions.
So first find the arrangements which have pink (P) and green (G) together:
[P,G] O <--- treat Pink and green as a group. P, G can be arranged in 2 ways within the group.

Now make a selection for the other card, O. This will be 7C1. Arrange these 3 cards: 2 * 7C1 * 2 = 28

Subtract 28 from the answer in (iii)

9P3 - 28 = 476 arrangements.

Appreciate it

 

[mohmed ahmed soliman]

part iii help maj june 2009 p6

 

[mohmed ahmed soliman]

help in oct nov 2003 part b II

 

[mohmed ahmed soliman]

oct nov 2004 `1 part ii help

 

[mohmed ahmed soliman]

help in part a nov 2009 part a p61

 

[Eugene99]

part iii helpView attachment 59790 maj june 2009 p6

i) No. of ways to chose a group: 13C10 x 12C9 x 6C4 x 7C4 = 33033000 or 33000000 (correct to 3 sf)

ii) No of ways of arranging 10 sopranos = 6P6 x 5P5 = 86400

iii) No of arrangement of 4 tenors and 4 basses with given condition: 2 ( 4P4 x 3P3) = 288

 

[Rizwan Javed]

part iii helpView attachment 59790 maj june 2009 p6

Let the Tenors be "T" and the Basses be "B".
The tenors are all together and so are the the bases.
TTT T BBBB

^The three tenors which can't stand with any Basses (B) are coloured red.
Now simply, arrange the basses (BBBB), this will be done in 4! ways. Then arrange the tenors (who can't stand with any basses) in the place shown, there will be 3! ways.
You can see that tenors ( TTT ) can either stand on the right or left like:
TTT T BBBB
OR
BBBB T TTT
So the total possible arrangements would be :

3! * 4! * 2 = 288

 

[Rizwan Javed]

help in oct nov 2003 part b IIView attachment 59791

Consider the 5 Boys as a group [BBBBB].
[BBBBB] GGG
Now simple do the arranging job. The 5 boys can be arranged with in the group in 5! ways. Once done with the arranging of the boys, arrange the 3 girls and the Boys (which are considered as a single unit). This will be done in 4! ways.
Combine all the arrangements, you'll get:

5! * 4! = 2880 arrangements.

 

[Rizwan Javed]

oct nov 2004 `1 part ii helpView attachment 59792

ARGENTINA
First separate the vowels and the consonants in this word:
vowel: AAEI
consonants: NNRGT

Let the vowels be represented by *
Let the consonants be represented by 'x'

The first one is to be a consonant, and then next one to be a vowel and so on. So the arrangement can be made this way:
x * x * x * x * x

Arrange the x's in the places shown above, keeping in mind there're 2 Ns (i.e. there's repetition!) : 5!/2!
Arrange the *s in the places shown above, keeping in mind there're 2 As (again there's repition!) : 4!/2!

Combining the results will give you:

5!/2! * 4!/2! = 720 arrangements.

 

[farhan141]

 

[mohmed ahmed soliman]

View attachment 59793 help in part a nov 2009 part a p61

help

 

[mohmed ahmed soliman]

help in part ii

 

[mohmed ahmed soliman]

help in (i)

 

[mohmed ahmed soliman]

help in V

 

[mohmed ahmed soliman]

help in II