# Help on three combinatorics questions?!

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!

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@CameronBui Could you show me how you edited the block quote above for future reference? Thanks! –  Billy Thorton Nov 1 '13 at 20:52
If you click the edit button, you can see exactly how the post is currently formatted. –  Cameron Buie Nov 1 '13 at 21:18

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.

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Thanks for your help! Dully noted about the multiple part questions! –  Billy Thorton Nov 1 '13 at 20:58

I see that Mr. Nicolas has posted an answer while I was working on mine. He has addressed the correctness of all parts of 1 and also of 2. I have altered my answer to address the correctness of 3, and also give alternative approaches to the portions of 1 that were not correct. I will also reiterate his encouragement to avoid posting multiple questions as one post in the future.

1a(ii): Let's start by assuming that we have some way to distinguish between each of the children, say by age. Now, there are $\frac{18!}{2!2!7!7!}$ ways to divide the toys up into numbered groups, with the first two groups having $2$ toys and the last two groups having $7$ toys. Also, there are $C(4,2)=\frac{4!}{2!2!}$ ways to determine which of the children will get $2$ toys and which of the children will get $7$ toys. Now, once we've grouped the toys and made the determination as to which children will get how many toys, we give the first group of toys to the older child who is to get two, the second to the younger such child, the third to the older child who is to get $7,$ and the fourth to the remaining child. Each distribution of toys can be obtained in this way. Consequently, $\frac{18!4!}{2!2!7!7!2!2!}$ is the answer.

1b(ii): We require each person to have at least one pear. Thus, the second case would be equivalent to the case of distributing 8 identical apples, 6 identical oranges, and 4 identical pears among 3 different people without restriction.

3a is incorrect. To start with, let's arrange the paired Fs, the L, the O, the R, the S, the paired Cs, and then N, in some order. There are $7!$ ways to do this. This gives us something like $$\text{_FF_L_O_R_S_CC_N_},$$ where the blanks are to be filled in with Es, or with nothing. Since we have 4 Es, and since they are to be nonadjacent, then we will be filling in exactly 4 of the 8 blanks with Es, and there are $C(8,4)$ ways to do this. Thus, there are $C(8,4)\cdot 7!$ ways to arrange the letters of EFFLORESCENCE with the given restrictions.

For 3b, the total is perfect, and your approach is fine, but you've miscalculated the number of ways an arrangement of WISCONSIN with no consecutive vowels can fail to meet the restrictions. So, to start with, we arrange the consonants, and there are indeed $\frac{6!}{2!2!}$ ways to do that. So, let's say we've got an arrangement of consonants like the following (where C represents a non-W consonant): $$\text{_W_C_C_C_C_C_}$$ In order to fail to meet the restrictions, both Is must be non-adjacent to W. One way to accomplish this is by having W not adjacent to any vowels, and there are $C(5,3)$ ways to accomplish this. The other way to accomplish this is by having W adjacent to O, but not to any I, and there are $C(2,1)\cdot C(5,2)$ ways to accomplish this. Hence, the number of ways an arrangement of WISCONSIN with no consecutive vowels can fail to meet the restrictions is $\frac{6!}{2!2!}\bigl(C(5,3)+C(2,1)C(5,2)\bigr).$

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Great answer! Thank you so much again, I really appreciate it! Do you think you could explain how you got C(2,1)*C(5,2) for the last part of 3(b)? I understand how you got C(2,1), but what about the C(5,2)? –  Billy Thorton Nov 1 '13 at 20:55
The $C(5,2)$ came about because we still have to place the $2$ Is, and we have $5$ spots to put them in (the $5$ that are non-adjacent to $W$). We then multiply $C(2,1)$ by $C(5,2)$ because our choice of a spot for the O has no effect on the available choices of spots for the Is. –  Cameron Buie Nov 1 '13 at 21:16