Do You Have Any Maths Question? Kindly Bring It On

    The number of 4 digit even numbers that can be formed using 0,1,2,3,4,5,6 without repetition is
A) 120
B) 300
C) 420
D) 20
E) 42


   RichyBlack, edusegzy, hollandis and others, please explain how

to solve this. I'm utterly confused. Would I ever understand permutation

and combination?

   
  @RichtBlack,

     Why can't I think this was when I see problems like

  this? All other math topics are like second nature to me

  except this one. Your logic is great RichyBlack. I now

 understand how to solve problems like this. But there's an

 issue here which I think will need some clarifications from

 you.

     Now look at this:

   

   Well, for S=0,  I understood how you assigned 6 choices to P,

                                                                               5 choices to Q,

                                                                        and 4 choices to R.

Now, for S= 2 , 4 , 6 :

      P  I think can't have 6 choices. Why?

  0 cannot start a number.

    With S=2,

  There will be 5 choices for P ( i.e, 1, 3, 4, 5, 6) excluding 0 because 0

 cannot be the first digit of a 4-digit number.

  Q will also have 5 choices ( i.e, 0, 3, 4, 5, 6) assuming the next digit assigned to Q was 1.

 R will have 4 choices then and S which is already = 2 will have just 1 choice.

 => There'll be 5*5*4= 100 numbers in E with S=2.

 Same applies for S=4 and S=6.

Extending the results,

 We find that there will be 120 + 100 + 100 +100 = 420 numbers in E.

 Answer = 420-- option C.

    Thanks man, RichyBlack for showing me the method to use.

  The problem I've got with questions like this is where to start from.

  I hope you'll continue to throw more light on questions I'll be posting.

  If I can get to understand this topic, I'll attribute it to this wonderful

 forum, edusegzy and you in particular for your insightful and vivid

 solutions. You kind of a math prodigy I think.

@Eyohimself

Try to identify the possible sets (or events) in the problem and their relationships, e.g. mutual exclusivity, independence, etc.

Further identify the particular set the problem is interested in.

Then use permutation, combination, etc. to count the elements in the set, if it's a counting problem (like the one just discussed).

As you solve more of these problems, you'll definitely get better at it.

 

Shells at the Shore
question

image with 3 buckets of 10 shells each and 4 loose shells image with 2 buckets of 10 shells each and 3 loose shells

image with 3 buckets of 10 shells each and 0 loose shells image with 2 buckets of 10 shells each and 1 loose shell

Jan, Kate, Mike, and Sam have lots of shells and many pails. They put groups of 10 shells in each pail.

    * Kate has 2 pails and 3 more shells left over.

    * Sam has 1 more pail than Kate, but no shells left over.

    * Jan has 4 more shells than Sam.

    * Kate has 2 more shells than Mike.

How many shells does each have?

  The wording of the question is kind of vaguely expressed.

  But I'll do it the way I understood.

  If Kate has 2 pails and 3 more shells leftover,

  => in the 2 pails there'll be 20 shells.

 In all, Kate has 20 + 3 = 23 shells then.

 Sam has 1 more pail than Kate,

 => Sam has 3 pails with a total of 30 shells.

 Jan has 30 + 4 = 34 shells.

 Kate has 2 more shells than Mike.

 Kate had 23 shells, => Mike will have 23 - 2 = 21 shells.

 In otherwords,
 
 Kate has 23 shells, Sam has 30 shells, Jan has 34 shells and Mike has 21 shells.

    I'm confused with the wording of the first proviso:

  Kate had (2 pails and 3 more shells) left over or

 he had 2 pails and (3 more shells) left over?

 The rationale to my solution is based on the latter.

 

  Can we solve this?

  A credit card number has 6 digits (between 1 to 9). The first two digits are 12 in that order, the third digit is bigger than 6, the forth is divisible by 3 and the fifth digit is 3 times the sixth. How many different credit card numbers exist?

   a)   27.
   b)   36.
   c)   72.
   d)   112.
   e)   422.

nice one richy blacky

i worked it  earlier and was getting 81 when i saw your solution i saw where i was getting it wrong

THanks man

   I'm interestd in how fast this problem could be solved and to see if I can find a faster approach.




The jewels in a crown consist of diamonds, rubies and emeralds. If the ratio of diamonds to rubies is 5/6

and the ratio of rubies to emeralds is 8/3, what is the least number of jewels that could be in the tiara?

a) 16

b) 22

c) 40

d) 53

e) 67

   

yes, decoupling the pair is a common mistake. You're welcome.


You're welcome man.

yes, I reside in Apapa, however, I'm currently visiting "the land of the free and home of the brave".

Pocket Change


  Your friend has a dollar bill and needs change for the vending machine. In your pocket, you have 7 pennies, 5 nickels, 8 dimes, and 3 quarters.

How many different ways can you make change for a dollar?

  He must sell it for :


  0.11*170270 + 0.055*170270 + 170270 = $198 364.55 ( $198 365--to the nearest $)

( Assuming they are making the gains on the cost price).

 

  Let d = dimes, n = nickels, q = quarters, p = pennies.

   quarter = 25 cent coin
 dime = a 10 cent coin
 nickel = 5 cent coin
 penny = 1 cent coin

  Lets start with d. Using d, the number of combination pairs possible are:

 (d,n)  (d,q,n)  (d,q,p)  (d,n,p)

 Without d, the number of possible combination pairs are:

 (n,p,q)   (n,q)

 This implies the total number of ways would be 11.

   A certain characteristic in a large population has a distribution that is symmetric about the mean m. If 68 percent of the distribution lies within one standard deviation d of the mean, what percent of the distribution is less than m + d ?
A. 16%B. 32%C. 48%D. 84%E. 92%

Given:

 A:B=3:2 => A/B = 3/2  => A= (3/2)B.

Lets assume amount invested in account A and account B are A and B respectively.

For account A,

at 15% ineterest rate, I'll see in my account when it matures an earning of:

 0.15*A

For B at 20%,you'll see:

0.2*B
 
 Now these two earnings total 18 000.

 => 0.2B + 0.15A = 18 000

 => 0.2B +  0.15*1.5B = 18 000

 Multiply through by 1000:

     200B + 225B = 18 000 000

=> B= (18 000 000)/425 = 42 352. 94

=> The amount invested in B was approximately 42 352

I'm assuming that you meant x^x (or X^X). If not, then you have to be clear with respect to what, because X is not the same as x.

Let y = x^x
Let log() represent the natural log function
Let exp() represent the exponential function

So, log(y) = x*log(x)

Let log(y) = P
So, y = exp(P) and P = x*log(x)

We know that dy/dx = dy/dp * dp/dx
We also know that dy/dp = exp(P)
And that dp/dx = x*(1/x) + log(x) = 1 + log(x)

Therefore, dy/dx = exp(P)*(1 + log(x)) = exp(x*log(x))*(1 + log(x)) = x^x*(1 + log(x))