Two questions on probability...?

Question

I have two questions.

1) In probability, when there is a question where order does not matter where replacement is allowed...there are two different formulas which the textbook talks about...

(n+r-1)C(r)

and

(n-1)C(r-1) + (n-1)C(r)

...so my question is: how do I know when to use which one of these formulas? What's the difference between them?

2) How many ways can you choose 3 times from the numbers 1 to 10 inclusive if order doesn't matter where these numbers are in non-decreasing order (i.e. order from smallest to highest with possible repitition) Answer= 12C3 = 220.

What I don't understand is...the formula n!/(r!(n-r)!) is used when order does not count and when replacement is not allowed...then why is this formula used here? Replacement is allowed in this question...

Thank you in advance

Answer 1

From Pascal's Triangle,

(n-1) C (r-1) + (n-1) C (r) = n C r

so the two formulas are equivalent.



Proof:

(n-1) C (r-1) + (n-1) C (r) =

(n-1)! / [ (r-1)! (n-1-r+1)! ] + (n-1)! / [ r! (n-1-r)! ] =

r (n-1)! / [ r (r-1)! (n-r)! ] + (n-1)! / [ r! (n-1-r)! ] =

r (n-1)! / [ r (r-1)! (n-r)! ] + (n-r) (n-1)! / [ r! (n-r) (n-1-r)! ] =

r (n-1)! / [ r (r-1)! (n-r)! ] + (n-r) (n-1)! / [ r! (n-r)! ] =

(r + n-r) (n-1)! / [ r! (n-r)! ] =

n! / (r! (n-r)! ) = nCr



You choose the one that is most convenient.



Now consider your problem of choosing from 1-10 three times, in nondecreasing order.

As soon as you choose a number, the remaining choices are limited by it.



Let N(a,r) = the same problem,

where a is the range 1 - a, and r is how many numbers you want.



Your problem is N(10, 3)



For the first number you can choose:

10 → N(1, 2) [ since now your choices are reduced to just 1 number, 10, and there are 2 picks left]

9 → N(2, 2) [ now for 2nd choice you have 2 choices, 9 OR 10, and 2 picks left]

8 → N(3, 2)

and so on

1 → N(10, 2)



So N(10,3) = sum of N(1,2) + N(2,2) + N(3,2) + N(4,2) + .... + N(10,2)



By the same process:

N(10, 2) = sum of N(1,1) + N(2,1) + N(3,1) + ... + N(10,1)



So N(a, r) = sum of (1,r-1) + N(2,r-1) + N(3, r-1) + .... + N(a, r-1)

and N(a, 1) = a [ since you have one pick with "a" choices]



So we have:

N[a,0] = 1 1 1 1 1   1   1  1 1 1 = a null case - just one way to choose "none"

N[a,1] = 1 2 3 4  5  6   7   8 9 10 = aC1 = first diagonal of Pascal's Triangle

N[a,2] = -- 1 3 6 10 21 28 36 45 55 = sum of all the N[a,1] to the left = (a+1)C2 = next diagonal

N[a,3] = -- -- 1 4 10 20 35 56 84 120 165 220 = (a+2)C3 = next diagonal

and N[a,r] = (a+r-1)Cr



Take a = 1 as the position where the first number appears in the row.



This is just Pascal's Triangle turned sideways:

the columns here are the normal rows, and the rows are the diagonals.



I don't know if that satisfies "why" or not, but it does show that it's the right thing to use.

Answer 2

perhaps you may like this less technical explanation for the 2 q's



q1

[see also "edit"]



nC(r) and (n-1)C(r-1) + (n-1)C(r) are identical. here's why

consider the r objects to be broken up into 2 piles of n-1 & 1

now you can either choose r-1 from 1st pile & 1 from the 2nd or all r from the 1st pile,

adding, you get (n-1)C(r-1) + (n-1)Cr = nCr



q2

this small q can be easily solved by summing up ways of choosing

3 of a kind, 2-1 of a kind and 1-1-1 of a kind to get 10c1 + 2*10c2+ 10c3 = 220

however, the better way is the formula given to you for combinations with repetitions

i've shortened #s for the purpose of explanation to ordered elements # 1-5

from which 4 are to be chosen

we can represent the permissible combinations using points and slashes.

note that we need only (5-1) slashes to make compartments for the 5 different #s



1,1,2,4 <=> • • / • / / • /



2,3,4,5 <=> / • / • / • / •



3,3,3,3 <=> / / • • • • / /



each string has 8 places with exactly 4 points & 4 slashes, & is unique for each combo

with each permissible combination, we can build a string putting exactly 4 points in 8 places, and fill the spare places with slashes.

total # of permissible combos = # of combos of dots+slashes together, = (5-1+4)C4



for your particular problem, therefore, it'll be (10-1+3)C3 = 12c3 ways



hope this has helped



edit:

------

in q1 , i see that (n+r-1)Cr with proper allocation of what n & r are is the standard "stars and bars" or "dots and slashes" or whatever approach for combos with repetitions, which has been used in q2.

Answer 3

a million. risk which you win at maximum as quickly as. = 5C0 (.2)^0 (.8)^5 + 5C1(.2)^a million(.8)^4 = a million*a million * (.8)^5 + 5* 0.2 * (.8)^4 = (.8)^4 ( .8 + 5*.2) = a million.8 * (.8)^4 = .73728 2. Pr (which you win all 5 circumstances) =5C5 *(.2)^5 *(.8)^0 = a million* (.2)^5 * a million = .00032 = a million/ 3125