₦airaland Forum

Bayur01:
Stop masturbation...you have probably masturbated before the sex

Masturbation sweet pass d real thing, why?

The Powerpuff Girls


Sisters Blossom, Bubbles and Buttercup, those babes can fight o.

There is another elegant approach to solving d numerator of this problem that is less tasking.

Interestingly, I had another guy solve it this way;

https://math.stackexchange.com/questions/3142455/the-draw-for-the-fifth-round-of-the-fa-cup

That was when I was still battling with the problem.

The problem scattered my brain sha, at d time.

Let's finish this motherfucker.

I have this set;

{4PO, [5!/(4! x 1!)] x 5PO, [6!/(4! x 2!)] x 6PO, [8!/(4! x 4!)] x 8PO}

But I need to eliminate repetitions and have something like this

{4PO, 5PO, 6PO, 8PO}

Where;

4PO stands for set of 8 team pairs with 4 correct only.

5PO stands for set of 8 team pairs with 5 correct only.

6PO stands for set of 8 team pairs with 6 correct only.

8PO stands for set of 8 team pairs with all 8 correct.
__________________________

This set;
{4PO, [5!/(4! x 1!)] x 5PO, [6!/(4! x 2!)] x 6PO, [8!/(4! x 4!)] x 8PO} = {4PO, 5 x 5PO, 15 x 6PO, 70 x 8PO}

has {8! / (4! x 4!)} x 7 x 5 x 3 x 1 = 7350 items.

I need to subtract 4 x 5PO to eliminate repetition and have only one 5PO.

Using
{8! / (5! x 3!)} x 5 x 3 x 1 = 840 ways

which is the set
{5PO, [6!/(5! x 1!)] x 6PO, [8!/(5! x 3!)] x 8PO} =

{5PO, 6 x 6PO, 56 x 8PO}

Multiplying with term [5!/(4! x 1!)] - 1 = 4

{4 x 5PO, 24 x 6PO, 4 x 56 x 8PO}


4 x 840 = 3360 ways

Need to subtract 3360 ways from 7350 = 3990

we have to add back these items 24 x 6PO and 4 x 56 x 8PO

For 24 x 6PO

24 x {8! / (6! x 2!)} x 3 x 1 = 2016 ways.

Which represents this set
24 x {6PO, 8!/(6! x 2!) x 8PO} = 24 x {6PO, 28 x 8PO}

Need to add 2016 ways to 3990 = 6006

Again, subtract these item 24 x 28 x 8PO

24 x 28 x 8!/(8! x 0!) = 672

Need to subtract 672 from 6006 = 5334

Add 4 x 56 x 8PO

4 x 56 x 8!/(8! x 0!) = 224

Need to add 224 to 5334 = 5558

Let's take stock, we are now at set {4PO, 5PO, 15 x 6PO, 70 x 8PO} with 5558 ways.

I need to subtract 14 x 6PO to eliminate repetition and have only one 6PO.

Using
{8! / (6! x 2!)} x 3 x 1 = 84 ways

which is the set
{6PO, [8!/(6! x 2!)] x 8PO} =

{6PO, 28 x 8PO}

Multiplying by 14

{14 x 6PO, 392 x 8PO}

= 14 x 84
= 1176

Need to subtract 1176 from 5558 = 4382

Add 392 x 1

Need to add 392 to 4382 = 4774


Let's take stock, we are now at set {4PO, 5PO, 6PO, 70 x 8PO} with 4774 ways.

I need to subtract 69 x 8PO to eliminate repetition and have only one 8PO.

Need to subtract 69 from 4774 = 4705

My guys, we have 4705 ways of arranging 8 team pairs with at least 4 correct choices. This can be represented as a set of subsets.

{4PO, 5PO, 6PO, 8PO}.

Now finally the probability of at least four correct guesses is

= 4705 / 2027025
= 0.0023211

Ok, let's proceed.

Why did I use d word set?

Let's assume I am given 3 letter: A, B and C. I want to form a set of two letter words, this set will contain the following 3 items.

AB
AC
BC

Where AB and BA is the same, the number of items in this set is 3. There is another way we can express this number, by using factorials.

Here, 3 combination 2, or 3! / (2! x 1!)

If we apply this same thinking to my problem, it starts to make sense.

{8! / (4! x 4!)} x 7 x 5 x 3 x 1

is the number of items in the set -- 8 team pairs with a minimum of 4 pairs being correct --

I know that 4 pairs are correct selections because I choose them from an imaginary set of correct pairs. I mean, assuming I cheated and found out all 8 correct pairs, I fit them in 4 boxes to make sure at the least 4 most be correct selections, there are 8! / (4! x 4!) ways this can be done.

It is going to get a little more involved from now on. If you are following pls acknowledge, I don't want to type d whole thing for nothing.

For 5 correct boxes only (5PO) we have something like this;

| A | B | C | D | E | x | y | n |

This combination will be counted more than once. Which we don't want because order is not important. Eg

| A | E | C | D | B | x | y | n |
| E | B | C | D | A | x | y | n |
| A | B | E | D | C | x | y | n |

They are actually counted 5 combination 4 times more than required.

So this big set of 8 pairs;
{8! / (4! x 4!)} x 7 x 5 x 3 x 1

Can be represented as a set of subsets.

{4PO, [5!/(4! x 1!)] x 5PO, [6!/(4! x 2!)] x 6PO, [8!/(4! x 4!)] x 8PO}

Where PO stands for correct pairs only. Again I chose 8PO instead of 7PO, they are d same.

Let me expand a little here.

I mentioned an imaginary set with all 8 pair correct. Let's say I tag these pairs with alphabets; A, B, C, D, E, F, G, H. For example, Man U vs Arsenal pair could be labeled A.

Then the four boxes can be fitted like this:

| A | B | C | D | z | x | y | n |

this is just one iteration out of 8! / (4! x 4!) .

The other boxes will contain any pair that we haven't placed in d previous 4 boxes.


{8! / (4! x 4!)} x 7 x 5 x 3 x 1

This expression counts every possible combination except less than 4 boxes being correct. Yes 6 boxes correct is in there, 7 boxes correct is also in there, only 4 boxes correct is also accounted for. It is a big set of various 8 pair combinations.

Let's split it into subsets.

1) 4 correct pairs only, rest incorrect.

2) 5 correct pairs only

3) 6 correct pairs only

4) 7 correct pairs only

We have 4 subsets, 8 correct pairs only is not counted because is the same with 7 correct, last two teams most be correct if all others are.

Really sad, no one is interested, I have similar thought provoking problems I would love to post.

Darivie04:
You said 3 days before you post solution na

I already had this part in my solution but I'm still trying to find the numerator.

Some tips:

A) We have 8 pairs, we can visualize it as 8 boxes, each holding a pair of teams. If we are asked to guess at least 4 correct then we know that 4 of these boxes will contain 4 correct pairs. The other boxes can contain any pair, correct or not correct.

B) Assuming there is a set of correct pairs, all 8 pairs correct. If we where to populate the 4 correct boxes in our previous reasoning with this set, we would have to match any 4 with our 4 boxes. This gives us 8 combination 4 possible ways;

8! / (4! * 4!).

C) The remaining 4 boxes can be fitted with any pair, correct or not. That is, we now have 8 teams to be entered into 4 boxes pairwise. Same reasoning as in my earlier post, this can be done in 7 x 5 x 3 x 1 ways.

Put everything together gives to give us;

{8! / (4! x 4!)} x 7 x 5 x 3 x 1

Sweet right? But there is a big problem, we counted some pairs more than once. Here come the beauty of set theory to our rescue.


To be continued ...

skitkid2:
Korean actors have better looking faces.

What type of silly comparison is this, are these two actors in the same age bracket? Why don't you post Brad Pitt's pix in his twenties and match up.



If I post my pix in my early 20's ibkayee go follow trip, now, not so much.

ifada123:
I am aware of that graphical method,
I would like if u can solve numerically

Let's use algebra then

(√2 -1)^x + (√2 +1)^x = 34

if (√2 -1)^x = a
then (√2 +1)^x = 34 - a

We can now find log of both sides of d equation.

x[log(√2 -1)] = log(a)
x[log(√2 +1)] = log(34 -a)

therefore,

log(a) / log(√2 -1) = log(34 -a) / log(√2 +1)

Reorganizing;

{log(√2 +1) / log(√2 -1)} * log(a) = log(34 -a)

then

-1 * log(a) = log(34 -a)

then, antilog

1/a = 34 - a

a^2 - 34a +1


Can you find a and then plug it back to find x?

talut:
You'll probably get bored once you approach the 3rd season

True, I enjoyed it till I started hating Clark and her Mother, they always win no matter what. Na them get all the luck, everything na them, men, I just tire.

But I have started watching it again sha, season six seems interesting. If am pleased I might watch season 5.

ifada123:
Hi,
Solve (√2 -1)^x + (√2+1)^x =34

This thread is for a specific problem, their are other place where you can post ur math questions.




Easiest solution is to plot (√2 -1)^x + (√2+1)^x = y

and find x at point y = 34

Darivie04:
You said 3 days before you post solution na

I already had this part in my solution but I'm still trying to find the numerator.

Sorry, I will post d other part in 3 days time.

Darivie04:
"You're not asked which pairs will be drawn at home..."

Could you explain this a bit?

This phrase is there to simplify the question, it means that for any selection, that is any pair, you don't need to bother your self with which team is home or away. For example - Man U vs Arsenal counts the same as Arsenal vs Man U.

Probability = Number of required outcomes / All possible outcomes.

Let's start with the denominator, it is much easier to deal with.

We have 16 teams and we want to pair off the teams, that is 8 pairs in all. To find the denominator we need to compute all possible pairs.

If we choose any team first we would have 15 choices to pair it with. After the first pair we are left with 14 teams. If we chose another team at random, we would have any of the 13 teams to pair it with. Continue like this for the remaining 6 pairs.

1st team to pair has 15 choices;
2nd team to pair has 13 choices;
3rd team to pair has 11 choices;
4th pair has 9 choices;
5th has 7;
6th has 5;
7th has 3;
8th team to pair has only 1 choice, the last team unpaired.

If you have n ways to do something and m ways to do another, then by logic we would have n x m ways to do both.

Therefore, we have 15 x 13 x 11 x 9 x 7 x 5 x 3x 1 ways in all.

This gives us 2027025, it is the total number of possible ways to pair these teams.

NNAMDIII:
your reps means nothing to me bro, take care

I will give out the answer in 3 days time. I guess I am the only one on NL who can solve this question.

Men, I am disappoint. Ok, does any one care for the answer?

.

You claim you know maths abi? You have a high IQ abi? Ok now, check this problem out:

https://www.nairaland.com/5121101/come-measure-iq

Where are the first class students, you think say everything na cramming, come and show your self if you are truly good.

Hint: The numerator is 4705, denominator is 2027025.

This is a counting problem, I suppose you know your combinatorics (combinations and permutations), probability and set theory well. Have fun!

The draw for the fifth round of the FA Cup is about to be made. There are 16 teams, leading to eight matches. Your task is to pair the teams off, in an attempt to guess as many as possible of the actual matches in the real Cup draw. You are not asked which teams will be drawn at home, just which pairs will be selected.

What is the probability of at least four correct guesses?

Gavrelino123:
This is good....
Study the last quote that i quoted you ...

freeman67:
Yes if you did not ask them to roll-over with interest. However, to be very sure of what has actually happened you have to contact them.

It's a rollover, confirmed from customer care, thank you.

OgogoroFreak:
Why not visit the bank to know what's going on?

I am off town. I have tried calling firstbank customer care, I will try again.

Gavrelino123:
Your 7.5 Million was Re-invested Automatically...
Possibly @ the rate 12.2% of 7,500,000=915,000
Less residual Bank charges=5k
That's why you are left with 910k..
This is a Fact..

Ok, I am sure I never ticked the re-invest option. So, for roll-over I only get to see my upfront interest, right?

Gurus in d house, I did t.bills of 7.5 mil last year to be due today. I just checked my phone and found out I was credited about 910k. I just don't understand whether I was short paid or it is on roll over, I have been doing this for some years now and I never roll over. Firstbank wants to deal with me o.

.

Tweetums:
I have 15m fixed deposit in Polaris bank but the return is very low every 91 days I receiving #336,575.. Pls, what are the processing to move to treasury bill... Thanks

It simple, find out from your account manager or any other staff in your bank if they buy treasury bills. If they don't then move your funds to other banks like Firstbank, GTB and Access that do.