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57) Given a collection of '2n' objects, 'n' idetical and the other 'n' all distinct, how many different subcollections of 'n' objects are there?

13) How many 8-digit sequences are there involving exactly six different digits?

Again the first solution gets all points.

13) How many 8-digit sequences are there involving exactly six different digits?

Again the first solution gets all points.

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Start your 7-day free trialAny collection of n will contain m form A and n - m from B

There are n!/n!(n-m)! ways of choosing m unordered objects from A and 1 way of chosing n-m from B

The total is is the sum over all possible m ie

sum (m=0 to n) n!/n!(n-m)!

this a well know summation and is 2^n

Given 6 digits then the number of 8 digit numbers is fom them is 6^8

there 8!/2!*6! ways of choosing the 6 digits giving a grand total of

6^8*8!/2!*6!

13) M things taken N at a time

Assuming order doesn't matter

M! / (N! * (M - N)!)

If order matters then

M! / (M - N)!

57) Sum of the binomial coeficients for (a +b) ^ N or 2^N in this case

mlmcc

but if the total is 3n (2n distincit and n similar)

tha naswer will be

sum sum(i=0,n,2n!/(2n-i)!)

so

0 - i may no choose any of n and all of m then the prem is m!/(m-n)!

1 - i may choose 1 from n and (n-1) of m then the prem is m!/(m-n-1)!

2- - i may choose 3 from n and (n-2) of m then the prem is m!/(m-n-2)!

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n- i may choose n from n and nothing of m then the prem is m!/(m-o)!=1

total is the sumetion

57) given p have n identical objects and q have n all distinct object

there is n+1 types of collection of n objects based on the number of object in p appear, there are :

1. n collection "without" p object ,( i assume order doesnt matter ) :

n C n = 1 subcollection

2. n collection "with 1" p object, so there is n-1 place left for n objects q :

n C n-1 = n subcollections

3. n collection "with 2" p object, so there is n-2 place left for n objects q :

n C n-2 = (n*n-1)/2 subcollections

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n-1. n collection "with n-2" p object, so there is 2 place left for n objects q :

n C 2 = (n*n-1)/2 subcollections

n. n collection "with n-1" p object, so there is 1 place left for n objects q :

n C 1 = n subcollections

n+1. n collection "with n" p object ,( i assume order doesnt matter ) :

n C 0 = 1 subcollection

total combination : nCn+nCn-1+...+nC1+nC0 = x

x is the total coefficient of (a+b)^n = ∑(nC(k1,k2))(a^k1)(b

if 'order is matter' we do in permutation way.

13) How many 8-digit sequences are there involving exactly six different digits?

Again the first solution gets all points.

# if order is not matter : there are 9 digits(the other one digit assume has put in 2 empty places ), so there is left 6 places with 9 available digits :

there are 9 C 6 = 84 digit sequences

# if order matter : there is (9 P 6)*(8 C 2(the way 2 same digit placed)) =14112

am i right?

hahan

Notice that each n-subset S of AUB is uniquely defined by its ``A-part'' which is some subset of A. So the total number of n-subsets S is equal to total number of all subsets of A, which is 2^n.

8) There two cases:

1) among 8 digits one occurs three times

number of sequences is 8!/3! = 6720

2) among 8 digits two occur two times each

number of sequences is 8!/(2!2!) = 10080

Total number of sequences is 6720+10080 = 16800

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The number involving more than 6 is 10*9*8*7*6*5*4 for the ordered 7 different digits which can be placed in 8 different ways (ie 8 spaces for the last digit to go into) and 10 choices for the last digit to go in

total=9*8*7*6*5*4*8*10 =10*9*8*(8*7*6*5*4)

This counts those with 7 and 8 different digits

For the number involving less than 6 (ie at least 4 the same) there 10 ways of choosing the repeated at least 4 times and 8*7*6*5/4! choices of how to arrange them in the 8 digit number. The remaing 4 positions can be filled with anything ie 10^4 possibilities

Giving a total of 10^4*10*(8*7*6*5)/4! = 10^5*(8*7*6*5*4)/3!

So not involving exactly 6 is

10*9*8*(8*7*6*5*4) + 10^5*(8*7*6*5*4)/3!

= 10*(8*7*6*5*4)(9*8 +10^4/3!)

Since there are 8^10 possible sequences then the total with exactly 6 is

8^10 - 10*(8*7*6*5*4)(9*8 +10^4/3!)