Some math ! With combinations !?
I'm supposed to believe you made these questions? I'll explain how to do #1: 1. A coin is tossed, and a standard number cube is rolled. What is the probability that the coin shows heads and the number cube shows an even number? When you toss the coin, there is 1/2 chance that it is heads. For the cube, there is a 3/6 = 1/2 chance it is even. Think about the sample space in which T maps to all possible numbers {1, 2, 3, ..., 6} and H maps to all possible numbers in the number set. There are 12 such elements in the sample space. Given that you want heads, the sample size cuts in half. Given that you want even numbers from those only, they cut in half again. So you only have a quarter of the sample size; 12(1/4) = 3. There is thus a 3/12 = 0.25 chance of the event happening. Consider using the multiplication rule now (since the events are independent, we may do this): P(H and Even) = P(H)P(Even) = (1/2)(1/2) = 1/4 = 0.25, the same answer as we expected. This corresponds to choice C.
If I am given 9 digits to work with, how many possible combinations of 2 digits are there?
how many different football matches can we see with 9 teams?36draw a 9x9 grid a match is a square, different are those in one of the halves, say "above" the diagonal. the diagonal is not valid. So you have 8 in the top row, 7 + 6 + 5 +...+1 = 36if you were familiar with combinatorials is just 9 over 2 = 9! / (9-2)! 2! = 36
Why does zero factorial (0!) equal one (1)?
This is a really interesting question. Back in my 10+2, most of the introductory books said that [math]0! = 1[/math] is by convention. But this "by convention" never ends the curiosity of anyone. While in my grads, I decided to give another try to find the answer. And then I came across a super amazing way of proving that [math]0! = 1[/math]. Here it goes..Before we proceed I think its good idea to define Factorial first, here is what Wikipedia says:It is a non-negative integer n, denoted by n!, is the product of all positive integer less than or equal to n.[math]n! = n*(n-1)*(n-2)*(n-3)*...3*2*1[/math]However the recursive definition of factorial is of more use in this proof.[math]n! =\begin{cases}1 & n = 0\\n*(n-1)! & n > 0\end{cases}[/math]Recursive definition of Factorial leads to one interesting way of expressing factorial numbers. [math]n! = \dfrac{(n+1)!}{(n+1)}[/math]This is valid since, as we expand [math](n+1)![/math] from recursive definition, we can cancel [math](n+1)[/math] term from both numerator and denominator to get [math]n![/math]. Or we can even calculate factorial in numerator and then evaluate the divisionFor example,[math]5! = \dfrac{6!}{6} = \dfrac{720}{6}[/math][math]4! = \dfrac{5!}{5} = \dfrac{120}{5}[/math][math]3! = \dfrac{4!}{4} = \dfrac{24}{4}[/math][math]2! = \dfrac{3!}{3} = \dfrac{6}{3}[/math][math]1! = \dfrac{2!}{2} = \dfrac{2}{2}[/math]In a similar way, if we try to express [math]0![/math] we get[math]0! = \dfrac{1!}{1} = 1[/math]And this ends our proof that [math]0!=1[/math].This proof is one of many ways, where [math]0![/math] leads to [math]1[/math]. But this one is quite explanatory in itself.There is one more interesting question I would like to share which was in my mind back in those days.WHY DO WE NEED [math]0![/math]?And the answer is computation of number of Combination.[math]{n \choose k} = \dfrac{n!}{k!(n-k)!}[/math]when [math]k = n[/math][math]{n \choose n} = \dfrac{n!}{n!*0!}[/math]This is one of many application of [math]0![/math]. But this example give a really good idea where we can make use of [math]0! = 1[/math].HTHCheers
How many zeros does 2008! or 2008 factorial end with?
Philo had the right idea, but he forgot to apply the same approach as with 10's to higher multiples of 5, like 25 and 125. There is no need whatsoever to consider multiples of 10 as being different to multiples of 5. Also, we can note that there are far more factors of 2 than there are of 5, i.e. the multiplicity of 5 is the limiting factor. (e.g. note that 100 is a multiple of 25, but so is 75. 1000 is a multiple of 125, but so is 625) So: there are no multiples of 3125 the 3 multiples of 625 supply 4 zeros each the (16-3) multiples of 125 supply 3 zeros each the (80-16) multiples of 25 supply 2 zeros each the (401-80) multiples of 5 supply 1 zero each SOLUTION: (3*4) + (13*3) + (64*2) + (321*1) = 12 + 39 + 128 + 321 = 500 Anyway this counting formula is more simply expressed if we simply counted them as: n = Σ (⌊2008/5⌋ + ⌊2008/5^2⌋+ ⌊2008/5^3⌋ + ⌊2008/5^4⌋ +...) where we continue until terms become zero i.e. 401 multiples of 5 supply 1 zero each 80 multiples of 5^2 supply another zero each 16 multiples of 5^3 supply another zero each 3 multiples of 5^4 supply another zero each again gives 500. And you can further note that: ⌊x/5^m⌋ = ⌊⌊x/5^(m-1)⌋/5⌋ i.e. just keep doing integer-divide-by-5, and summing the partial counts of factors till they reach 0. You could do it in one for-loop in C language: int x=2008, tens=0, factors; for (factors=x/5; factors>0; factors/=5) { tens = tens + factors; } printf("%d! -> %d factors of ten\n", x, tens); [Philo undercounted the higher multiplicities of 5 on this statement: "finally, the 201 numbers that end in 5 will provide [only one] 0 when multiplied by an even number"] SECOND PART: Suppose 2008 is written in a base other than 10 (say 8). Is it possible to solve it without first converting the number back to the decimal system (or base 10 system)? Yes it is. (Just the divisibility test won't look very intuitive.) The code I wrote above is totally invariant of base, you would only need to know the base when doing input or output.
Permutations and Combinations: In how many ways can we divide n distinct objects into r distinct groups of equal or unequal sizes having at least 1 object each?
This problem can be solved in 2 ways.Method 1:Solution to this problem is the same as finding solutions to the equation.[math]x_{1} + x_{2} + ... + x_{r} = n[/math]Now, solving the above is easy using binomial theorem i.e.Coefficient of [math]x^{n} in (x+x^{2} + x^{3} + ....)^{r}[/math]= Coefficient of [math]x^{n} in x^{r}(1+x + x^2 + ....)^{r}[/math]= Coefficient of [math]x^{n-r} in (1+x+x^2 + ....)^{r}[/math]But [math]1+x+x^2 + .... = (1-x)^{-1}[/math]Expanding [math](1+x +x^2 + ...)^{r} = (1-x)^{-r}[/math][math](1 + \binom{r}{1}.x + \binom{r+1}{2}.x^2 + .... \binom{n-1}{n-r}.x^{n-r} + ..)[/math]SO required answer is [math]\binom{n-1}{n-r} = \binom{n-1}{r-1}[/math] .Method 2:The other method is a pictorial one.You can think of the circles as the n objects. The vertical straight lines can be thought of as the symbols to partition the n objects into r groups. So, there will be n circles and (r-1) vertical straight lines.Now there are n - 1 such places(Since there is no line for first or last position, as there is a group of at least size one).So our job is to select (r-1) lines from (n-1) positions.So, required answer = [math]\binom{n-1}{r-1}[/math]