I was having trouble on these combinatorics questions, would anyone be able to help?
(P.S. I'm new to this forum, so bear with the editing – I know it sucks, but I tried to make it as readable as possible)
1.)
a) How many ways are there to distribute 18 different toys among four children…
i) Without restrictions
ii) If two children get seven toys and two children get two toys?
b) How many ways are there to distribute eight (identical) apples, six (identical) oranges, and seven(identical) pears among three different people…
i) Without restrictions?
ii) With each person getting at least one pear?
2) Suppose that 30 different computer games and 20 different toys are to be distributed among three different bags of Christmas presents. The first bag is to have 20 of the computer games. The second bag is to have 15 toys. The third bag is to have 15 presents, any mixture of games and toys. How many ways are there to distribute these 50 presents among the three bags?
3)
a) What fraction of all arrangements of EFFLORESCENCE has consecutive Cs and consecutive Fs, but no consecutive Es?
b) Among all arrangements of WISCONSIN without any pair of consecutive vowels, what fraction have W adjacent to an I?
My attempts:
1) a) i) 4^18
ii) 18!/(2!*2!)
b) i) C(8+3-1,8)C(6+3-1,6)(7+3-1,7) = C(10,8)*C(8,6)*C(9,6)
ii) C(8+3-1,8)C(6+3-1,6)(7*6*5)*C(4+3-1,4) = C(10,8)*C(8,6)*210*C(6,4)
2) (30 nPr 20)*(20 nPr 15)
3) a) total = 13!/(4!2!2!)
with restrictions = C(3+5-1,3)*6! = C(7,3)6!
fraction = with restrictions/total = 1/2574
b) total = (3!/(2!1!))(6!/(2!2!))C(7,3)
with restrictions = (6!/(2!2!))(C(5,3)+C(2,1))
answer = total – with restrictions
I'm not sure that I'm doing it correctly. I can explain any of my answers as needed. Detailed and good explained answers would be very much appreciated! Thank you in advanced for your help, Its greatly appreciated!
Best Answer
We make a few comments. 1(a)(i) is right. For 1(a)(ii), I would choose the two kids who get $7$ toys each. That can be done in $\binom{4}{2}$ ways. The toys for the shorter of these two kids can be chosen in $\binom{18}{7}$ ways, then the toys for the taller one can be chosen in $\binom{11}{7}$ ways, then the toys for the shorter of the two remaining kids can be chosen in $\binom{4}{2}$ ways, for a total of $\binom{4}{2}\binom{18}{11}\binom{11}{7}\binom{4}{2}$.
The question 1(b) uses standard Stars and Bars. 1(b)(i) is attacked correctly. But 1(b)(ii) is quite wrong, The number of ways to distribute the pears so that each gets at least one is $\binom{6}{2}$.
Question 2 is right apart from notational issues, you should write ${}_{30}P_{20}$, for example.
Remark: It is not a good idea to ask a many parts question on MSE, particularly if the parts are not closely related.