first exercise With how many was we can put 20 same CD, in 3 slots?One slot takes 10 CD the most.There must not be ,a slot that will be empty.
My answer was
Because it is same that means i have two typology C(n,r)=n!/r!(n-r)! and C(n+r-1,r)=(n+r-1)!/r!(n-1).
I have use the second
- C(n+r-1,r)=(n+r-1)!/r!(n-1) so i use n=20 and r =3 and it is C(20+3-1,3)=(20+3-1)!/3!(20-1). am i right?
second exercise
We give digits 1,2,4,5,6,7,9.How many even numbers,four-digits,bigger than 5000,can we do with those digits,with not use the same digit more than 1 time? My answer is I have the P(n,r)=n!/(n-r)! i didn't do anything cause i confuse who is n and who is r.
I have big problem on this ,i would like good explanation to understand and if it is possible some links to understand it more.I have seen many examples but i cant progress to solve.Thanks in advance
The second exercise is far more fundamental than those exercises that require the use of permutations or combinations. It relies on the tool of The Rule of Product and the Rule of Sum.
Paraphrased, the Rule of Product states that if you want to count how many ways there are to accomplish a task and each possible outcome can be uniquely described via a sequence of steps such that the number of options for a step does not rely on the previously made choices (though the options themselves may change) then the total number of outcomes is equal to the product of the number of choices for each step.
Paraphrased, the Rule of Sum states that if you want to count the number of outcomes of a scenario and you can break apart the outcomes into separate non-overlapping groups, you may count the amounts within each group individually and then add the results together.
Here, break up using the cases:
The first digit is odd
The first digit is even
Next, break up each case into the following steps:
Pick the first digit. It must be a 5, 6, 7, 8 or 9. There are 3 choices here if in the first case where the first digit is odd and 2 choices if in the second case.
Pick the fourth digit. It cannot be what was chosen for the first digit and must be an even number. If in the first case there will be 4 options. If in the second case there will be only 3 options.
Pick the second digit. It cannot be what was chosen for the first digit or fourth digit but may be anything else from those available. There are 7 choices available here.
Pick the third digit. It cannot be what was chosen for any of the earlier steps. There are 7 remaining choices available.
Multiplying the number of options available for each step and adding over our cases we get our total number of outcomes as $3\times 4\times 7\times 6 + 2\times 3\times 7\times 6$.
The rule of product is a terribly useful tool for counting problems and should be one of the first things you ever try to think about using in many elementary scenarios. Even the formula for permutations comes as a direct application of the rule of product.
As for the first problem, this is unfortunately rather tedious.
Your attempted answer of $\binom{20+3-1}{3\color{red}{-1}}$ (you forgot the -1 at the bottom there) was incorrect since it completely ignores the condition that a slot can hold a maximum of 10 cds. It would have been correct if not for that (after including the missing -1).
The easiest solution would be to use a generating function. The answer would be the coefficient of the $x^{20}$ term of the expansion of $(1+x+x^2+\dots+x^{10})^3$ which would be $66$. This solution however requires you have a computer algebra system capable of expanding the product or requires you be satisfied with not computing the actual numerical value.
You could also do this a bit more tediously by inclusion-exclusion based on the events where one (or more (though more is impossible in this example)) of the slots has too many cds in it.
This would give $\binom{22}{2} - 3\binom{11}{2}+0-0 = 66$
As for something to help you remember when you should or shouldn't have the extra $-1$ on the bottom, see my answer here.