MATH TEASER - Page 2

Assuming 1 and 10 are 2 different digits(?)

we can have 362880 numbers with the digits from 1 to 10, no number occupying its correct position

9*8*7*6*5*4*3*2 =362880

If you say 10 also cannot appear in first place because it has a '1', then we will hv 322560 nos.

Let me know if this is correct! 😊

awww Giri, munnadiye pathuttiya?? was busy with
kolangal update as Niti is sick -

9 X 8 x 7 x 6 x 5 x 4 x 3 x 2 x10 = 3628800 - however
u change will still comply with the rule !!
- sariya Tamizha???

both of you are wrong! the correct answer is not 3628800 - i'll post the answer very soon!

I think its 1 334 961 ways...right???

meli your answer is correct! I will explain the answer to this question soon! tried wriiting but it's confusing me ... but i'll post an explanation soon!

Please post more maths questions ... i can't live without doing maths riddles!

heres another one... well, this is kinda maths teaser... there are 4 children and they all have to cross the road... lets name the children
child 1 - A
child 2 - B
child 3 - C
child 4 - d

child A takes 10seconds to cross the road, child B takes 5seconds, child C takes 2seconds and child D only takes 1second (he's super fast)! and as it is night time they have to take a torch with them - they only have 1 torch! and they can't all cross at the same time only 2 at a time. however, if A and B are crossing the road then it'll take 10seconds (the longest time will be taken as the faster one [5sec] has to slow down and walk slower with the slow one)... can you get all of them across the road in 17secs or less?

Originally posted by: ~*Thamizhan*~

Another teaser!
you should list the numbers 1 through 10 so that no number appears in it's own position... how many ways are there? (eg. 1 should not be 1st in the list, 2 should not be second in the list and so on)...

Well, this is a very complicated question. I would like to start with a slightly easier question, so that you can get used to the type of reasoning I will be using to explain how I solved this problem.

To start with, consider this warm-up problem: How many ways are there to arrange the numbers 1 through 10 without any restrictions?

We can answer this question by taking the numbers one at a time. For example, there are 10 places we could put the -1- in the sequence: it could go in any of the 10 places.

Now for EACH of those 10 places, how many places are there where you could put the -2-? It could go anywhere except in the place that is already taken by the -1-, so there are 9 places the -2- could go for each of the 10 places that the -1- can go. That means that there are 10x9 = 90 different ways to put the -1- and the -2- in the ten places in the sequence.

Next, for EACH of these 90 places, how many places could you put the -3-? I think you can see there will be 8 places the -3- can go (anywhere except wherever the -1- is and wherever the -2- is) so now we have 10x9x8 = 720 different ways to put the -1- and the -2- and the -3- in the ten places in the sequence.

By similar logic, for EACH of the 720 ways to arrange the first 3 digits, there will be 7 places to put the -4-, and 10x9x8x7 = 5,040 ways to arrange the first 4 digits.

If you continue this process for all 10 numbers, you will see that the answer to this warm-up problem is 10x9x8x7x6x5x4x3x2x1=3,628,800. (This expression is called ten factorial, and is often notated as 10!) You may notice a pattern developing before you actualy complete all
of the steps, which often helps find a sort of short-cut to the
answer.

Now let us tackle the real problem you asked:

How many places are there where you could put the -1- ? I suppose there are 9 places you could put it, since it can go anywhere except the first place.

Okay. Now, for EACH of those 9 places, how many ways are there to arrange the other numbers? Lets start with the -2- . If the -1- were in the -2-s place, then the -2- could go anywhere else, so there would be 9 places. But, if the -1- were anywhere else the -2- could not go in its own place or in the place where the -1- was, so then there are only 8 places for the -2- to go. So how many ways are there to put both the -1- and the -2- into the 10 places (in position), without regard to the other numbers? 9 ways with the -1- in the -2-s place plus 8 ways for EACH of the other 8 places that the -1- can go.
9+8x8 = 73.

Now let's find out how many ways there are to put the -3- into the picture for EACH of the 73 ways we have found so far. Well, if the -1- or the -2- were in the -3-s place, then there would be 8 places that the -3- could go, right? But if neither the -1- nor the -2- were in the -3-s place, then the -3- would only have 7 places it could go.So, how many of the 73 ways to put the -1- and the -2- in position would have either the -1- or the -2- in the -3-s place? There are only 8 ways to put these two numbers in position which have the -1- in the -3-s place, and there are also 8 ways to put these two numbers in place which have the -2- in the -3-s place, so there are only 16 ways
with one of these numbers in the -3-s place.

The other 57 ways to put the -1- and the -2- in position all have the -3-s place left open. (73-16=57) So (16 possible arrangements of -1- and -2- with the -3-s place occupied) x (8 places for the -3- to go if the -3-s place is occupied) + (57 possible arrangements of -1- and -2- with the -3-s place vacant) x (7 places for the -3- to go if the -3-s place is vacant) = 16x8+57x7 = 527 ways to put the first 3 numbers in
position.

We are trying to look for a pattern so that we do not have to do all 10 numbers this way.

Lets do -4-. For each of the 527 possible ways to put the first 3 numbers in position, how many places are there that the 4 could go? By using similar logic to the steps above, in 171 of those 527 cases the -4-s place will be occupied, and in the other 356 cases it will not be, so there are 171x7+356x6= 3,333 ways to put the first 4 numbers in position.

Now I have begun to notice a pattern emerging. Each time we add a new digit to the problem, the first question we ask is: How many of the possible arrangements from the previous step will leave the new digits own place open? But that is really a similar problem to the original question, except with less possible places to put the numbers instead of 10. So I noticed that the answer to this question is related to the answer from the previous step. I know that this is very confusing at this point, and I don't expect your class to be able to follow all of it, but I wanted to show you what I did, rather than just tell you the answer. Let me show you what I mean:

number of equation number of ways to
digits arrange these digits

1   ; ; 0 x 10 + 1 x 9 = 9

2   ; ; 1 x 9 + 8 x 8 = 73

3   ; ; 16 x 8 + 57 x 7 = 527
(2x8) (73-16)

4   ; ; 171 x 7 + 356 x 6 = 3,333
(3x57) (527-171)

5 1,424 x 6 + 1,909 x 5 = 18,089
(4x356) (3,333-1,424)

6 9,545 x 5 + 8,544 x 4 = 81,901
(5x1,909) (18,089-9,545)

7 51,264 x 4 + 30,637 x 3 = 269,967

Do you see the pattern? I used a spreadsheet to continue this
iterative process, and for all 10 digits the answer is: 1,334,961

These possible arrangements are called permutations, which is an important concept in the field of probability. There is probably a formula we could have looked up to tell us the answer, but sometimes its more fun (and more educational) to figure it out for yourself.

Explanation by Doctor Barney!

Originally posted by: ~*Thamizhan*~

meli your answer is correct! I will explain the answer to this question soon! tried wriiting but it's confusing me ... but i'll post an explanation soon!

Please post more maths questions ... i can't live without doing maths riddles!

heres another one... well, this is kinda maths teaser... there are 4 children and they all have to cross the road... lets name the children
child 1 - A
child 2 - B
child 3 - C
child 4 - d

child A takes 10seconds to cross the road, child B takes 5seconds, child C takes 2seconds and child D only takes 1second (he's super fast)! and as it is night time they have to take a torch with them - they only have 1 torch! and they can't all cross at the same time only 2 at a time. however, if A and B are crossing the road then it'll take 10seconds (the longest time will be taken as the faster one [5sec] has to slow down and walk slower with the slow one)... can you get all of them across the road in 17secs or less?

Then they can cross in 17 seconds

C(2) & D (1) = 2 seconds

D returns back = 1 second

B(5) and A(10) = 10 seconds

C returns back=2 seconds

C & D = 2 seconds

Total 17 seconds

how abot they can cross in 14 seconds.....

first A(10) and B(5) =10 seconds.


D(1) to get torch =1 second.

D returns back =1 second.

C(2) and D(1) = 2 seconds.

total 14 seconds....

But D cannot travel across the road without the torch rathy......so i guess 17 is the only possible answer to that question....

Welcome ginny 😊 ,

I am sure you will have a wonderful time in IF.