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According to my book:

1.The fundamental principle of counting is used to count the number of possible ways in which a task can be done without actually counting manually.

2.Under the fundamental principle of counting comes the principle of addition and the principle of multiplication.

3.The principle of addition states if a one task can be one done in $m$ ways and another task which is MUTUALLY EXCLUSIVE of the first task can be done in $n$ ways, then the number of possible ways in which either can be done is $m+n$.This result can be generalized.

4.The principle of multiplication states that if one task can be done in $m$ ways and another task which is INDEPENDENT of the first task can be done in $n$ ways, after the first task has been performed, then the number of possible ways in which both the tasks can be done is $m \times n$.This result can be generalized.

5.Permutations refer to the possible arrangements of a collection of objects.

6.Combinations refer to the possible selections of a collection of objects.

My problems:

1.I was having a lot of problems understanding the difference between the principle of addition and the principle of multiplication. I thought about it a lot and this is my interpretation:

(a).The addition principle is applied when we want to calculate the number of possible ways to perform a task (perform any one of the subtasks).

The task consists of $k$ subtasks ($n_1, n_2, n_3, ..., n_k$), the subtasks can be done in $m_1, m_2, m_3, ..., m_k$ ways respectively. Thus the number of possible ways to do the task is $m_1 + m_2 + m_3 +...+ m_k$.

(b).The multiplication principle is applied when we want to calculate the number of possible ways to perform a task(perform all the subtasks).The task consists of $k$ subtasks($n_1, n_2, n_3, ..., n_k$), the subtasks can be done in $m_1, m_2, m_3, ..., m_k$ ways respectively.Thus the number of possible ways to do the task is $m_1 \times m_2 \times m_3 \times ... \times m_k$.

Am I correct?

2.I cannot understand the significance of the word MUTUALLY EXCLUSIVE in point number 3 and the word INDEPENDENT in point number 4. What does it mean when we say that one task is independent of the other and what is its significance here.

3.What is the difference between principle of multiplication and permutation. I guess permutation is a special case of the principle of multiplication.

4.What is the difference between principle of addition and combination.

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  1. Yes, you are correct. A more down to earth example, if I have $4$ apples and $3$ oranges in a basket, there are $3+4=7$ ways I can pick one fruit from the basket, and there are $3\cdot 4$ ways I can pick one apple and one orange from the basket.

  2. The independent means that no matter how one task is performed, the number of ways you can perform the second task is the same. For example, the tasks "pick one apple" and "pick one orange" are independent in the previous example, since no matter what apple I pick, I still have the same $3$ oranges to pick from.$$$$ On the other hand, if my first "task" is "pick one fruit" and the second is "pick one orange", then the number of ways I can perform the second task depends on the way I performed my first task. If I picked an apple, I have $3$ ways of performing the second task. If I picked an orange, I have only $2$.

  3. The formula for the number of permutations is obtained by applying the principle of multiplication.

  4. These are two different things, the principle of addition is a general principle, while combinations are specific collections of objects from a set.

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  • $\begingroup$ In point number 3, by principle of permutation you mean principle of multiplication.Right? $\endgroup$ – MrAP Nov 24 '16 at 10:43
  • $\begingroup$ @MrAP of course, yes. Sorry. $\endgroup$ – 5xum Nov 24 '16 at 10:44
  • $\begingroup$ Sorry for extending this conversation but in point number 2, what if the task is to pick two fruits? In how many ways can it be done? $\endgroup$ – MrAP Jul 22 '17 at 22:21
  • $\begingroup$ @MrAP If it's just two fruits, then you are looking for the number of combinations (so, $n\choose k$ is the answer) $\endgroup$ – 5xum Jul 24 '17 at 6:53
  • $\begingroup$ I am sorry to extend this conversation again. I have edited my question a little bit. Please see the edited question. $\endgroup$ – MrAP Oct 3 '17 at 15:02

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