Question 1182241: Hello, I just need someone to check my answers for these questions. Thank You.
1. In how many different ways can seven books be arranged on a shelf of two particular books must always be next to each other?
for this I did 5! * 2! * 6 (5! from the 5 left over books and 2! from the two books, 6 positions for the girls to start with)
2. How many ways can five pieces of candy be divided between two children if each child receives at least one piece of candy?
There are two cases for this question.
1st case:
One child gets one candy, and they other one gets the rest. So
I did 5 choose 1 times 5 choose 4 which is 25 ways.
2nd case:
one child get 2 candies, while the other one get 3.
So I did 5 choose 2 times 5 choose 3, which is 100 ways.
Finally I add these values up 100 + 25 = 125
2. A team of five is to be chosen from a group of eight students. The two tallest students cannot both be on the team. How many different teams can be formed?
for this one, I did 8 choose 5 minus 6 choose 3 (because the two tallest are removed), which equals t0 56 - 20 = 36.
3. A group of 8 people consists of 6 adults and two children. Calculate the number of ways in which five people can be selected from these 8 in each of the cases:
(a)order of selection is unimportant
(b)the five people are selected in a definite order (I don't understand this one)
(c)order of selection is unimportant and at most one child may be selected.
for (a) I did 8 choose 5. 56 ways.
(b) I don't understand b at all
(c) I did 7 choose 4 because, one child is guaranteed to be selected. 35 ways.
Answer by ikleyn(52782)   (Show Source):
You can put this solution on YOUR website! .
In this my post, I will solve and explain you ONLY ONE problem of several posted.
It is the problem n.2 about candies.
In how many ways can five pieces of candy be divided between
two children if each child receives at least one piece of candy?
In this problem, candies are considered as NON-DISTINGUISHABLE - - - so the only difference
between different cases is in the NUMBER of candies each child get.
Without going into high and complicated Math, we can write the FULL LIST of all possible solutions/answers
T A B L E
1st child 2nd child
1 4
2 3
3 2
4 1
It is the set of all possible different solutions.
ANSWER. There are 4 (four) different solutions according to 4 different lines in the Table.
Solved.
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Post your problems SEPARATELY.
It is the rule and the policy of this forum.
Each your problem/question requires to give you a small lesson/lecture, and
if some tutor will answer everything in one piece, you will get a mess.
It is why the requirement "one problem per post" is established.
Come again and enjoy my teaching ( ! )
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By the way, if the candies are considered as DISTINGUISHABLE in this problem, then the answer is 30:
there are 30 different solutions in this case.
30 = 32 -2; 32 =  : there are different subsets of the set of 5 candies
(which the 1st child get; and then the 2nd child gets the rest just without a choice).
And we subtract 2 from 32 to exclude the degenerated (and prohibited) cases when one child gets 0 (zero) or all 5 candies.
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