Solve only problem 3.2.2 So the number of ways the presents can be laid out (giv
ID: 2965979 • Letter: S
Question
Solve only problem 3.2.2
So the number of ways the presents can be laid out (given the distribution of the presents to the children) is a product of factorials: Thus the number of ways of distributing the presents is We can describe the procedure of distributing the presents as follows. First, we select n1 presents and give them to the first child. This can be done in ways. Then we select n2 presents from the remaining n - n1 and give them to the second child, etc. Complete this argument and show that it leads to the same result as the previous one. The following special cases should be familiar from previous problems and theorems. Explain why.Explanation / Answer
No. of ways of distributing the n1 gift from n gifts to 1 child = nCn1 = n! / n1! (n - n1)!
No. of ways of distributing the n2 gift from n-n1 gifts to 2nd child = n-n1Cn2 = (n-n1)! / n2! (n - n1 - n2)!
Similarly ,
No. of ways of distributing the nk-1 gift from n-n1 - n2 ...- nk-1 gifts to kth child
= n-n1-n2.....-nk-1Cnk = (n-n1-n2.....-nk-1)! / nk! (n-n1-n2.....-nk-1-nk)!
Hence total ways = nCn1 * n-n1Cn2 * ... * n-n1-n2.....-nk-1Cnk * .....* mCm
= [ n! / n1! (n - n1)! ] * [ (n-n1)! / n2! (n - n1 - n2)!] * .... *[(n-n1-n2.....-nk-1)! / nk! (n-n1-n2.....-nk-1-nk)! ]* ...*m!C0!m!
= [ n! / n1! (n - n1)! ]* [(n-n1)! / n2! (n - n1 - n2)! ] * .... *[ (n-n1-n2.....-nk-1)! / nk! (n-n1-n2.....-nk-1-nk)! ]...* [m! / m!0! ]
Cancelling n-n1-n2.....-nk-1 term from denominator of the previous with the numerator of the next term we get
= n! / ( n1! n2! n3! ..... nk! )
2.)
a.)
If n = k and n1=n2 = .... = k
Then we are distributing k gifts among k people implying 1 gift per persons and as it is apparent the number of ways has to be k!
By using n! / ( n1! n2! n3! ..... nk! ) = k! / ( 1! 1! 1! ..... 1! ) = k!
hence we also get the same answer by the formula.
b.)
Now we are giving 1 gift to each k-1 persons and rest left out gifts to the kth person
Hence total ways =(No of ways of distributing k-1 gifts among k-1 people with 1 gift per persons) *(No. of ways of choosing k-1 gifts from n gifts)*(No of ways of giving the left out gift to the person left)
= ( k-1Ck-1(k-1)! )*(nCn-k+1)*(1)
= (k-1)!*(n! / (k-1)!(n-k+1)! )
= n! /(n-k+1)!
By using n! / ( n1! n2! n3! ..... nk! ) = n! / ( 1! 1! 1! ..... (n-k+1)! ) =n! /(n-k+1)!
c.)
For k =2 ,
No. of ways of distributing the n1 gift from n gifts to 1 child = nCn1 = n! / n1! (n - n1)!
No. of ways of distributing the remaining n2( = n-n1) gifts to 2nd child = 1
Total ways = nCn1 = (n!) / n1! *n-n1! = n! / n1! n2!
By using n! / ( n1! n2! n3! ..... nk! ) = n! / n1! n2!
d.)
k = 3 , n=6
Here we try to divide 2 gifts equally among 6 people.
Hence, ways of distributing 2 gifts to first person = 6! / 4!*2!
ways of distributing 2 gifts from 4 gifts remianing to second person = 4! / 2!*2!
ways of distributing 2 gifts from 2 gifts remianing to 3rd person = 1
Total ways = (6! / 4!*2! ) * (4! / 2!*2! ) * 1 = 6! / (2!*!2!*2! )
By using n! / ( n1! n2! n3! ..... nk! ) = 6! / ( 2! 2! 2! )
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