I am confused by the following exercise. First, I will present the exercise with solution, after that there is a problem that I have about this exercise.

Exercise. From 7 consonants and 5 vowels, how many words can be formed consisting of 4 different consonants and 3 different vowels? The words need not have meaning.

Solution. The 4 different consonants can be selected in $_{7}C_{4}$ ways, the 3 different vowels can be selected in $_{5}C_{3}$ ways, and the resulting 7 different letters (4 consonants, 3 vowels) can then be arranged among themselves in $_{7}P_{7}$ - $7!$ ways. Then Number of words - $_{7}C_{4}$ $_{5}C_{3} 7! = 1,764,000$.

Problem. The problem is I don't understand when I need to use combinations and when permutation and this examples demonstrates my confusion. The idea of the solution is to generate the different sets of letters (combinations) of two kinds all possible sets of 4 consonants from 7 consonants and sets of 3 vowels from 5 vowels. After generation, mix up everything by multiplying by 7!. Without learning the solution I thought that the permutation is more approprite for this case and can avoid the 3rd step - mixing up. So in my opinion something like $_{7}P_{4}$$_{5}P_{3}$ should give the right answer, however here is the second confusion we should somehow to mix up two these sets together.

Of course, the solution with permutations is not right, but it seems like the exercise can be solved by using permutations.

Can you solve this exercise with permutations?

Can you give an explanations when is better to use combinations and when permutations.


2 Answers 2


You should think of this as a two-part problem: first you must pick the $7$ letters to be used, and then you must arrange them. You can’t arrange them until you’ve picked them. Picking them is just picking sets of things: order is irrelevant, so you’re counting combinations. Arranging them, on the other hand, is clearly a matter os specifying an order, so you’re dealing with permutations.

Are there other ways to solve the problem? Yes, but they’re more difficult. You could begin by picking an ordered string of $4$ consonants; this can be done, as you said, in $_7P_4$ ways. You now have a skeleton $s_1C_1s_2C_2s_3C_3s_4C_4s_5$, where $C_1,C_2,C_3$, and $C_4$ are the consonants, and $s_1,s_2,s_3,s_4$, and $s_5$ are the slots into which you can insert vowels. There are now $_5P_3$ ways to select an ordered string $V_1V_2V_3$ of $3$ vowels, and the problem is to count the ways to fit these vowels into the $5$ open slots in the consonant skeleton. Doing that is a matter of selecting a multiset of $3$ not necessarily distinct slots from the $5$ available. This can be done in


ways, so the there are $840\cdot60\cdot35=1,764,000$ such words, exactly the figure obtained by the other computation.

  • $\begingroup$ Thank you very much for the great explanation, I wish this was a way of teaching in school, the feeling is that the things are interconnected (combination and permutation) and it should be mentioned like in your answer, even thought one of the ways is more difficult than the other. $\endgroup$
    – user16168
    Sep 11, 2013 at 10:34
  • $\begingroup$ Second part of "doing other wise" is simply amazing. $\endgroup$ Sep 11, 2013 at 10:39
  • $\begingroup$ @user16168: You’re very welcome. $\endgroup$ Sep 11, 2013 at 11:27

By saying $7P4 \times 5P3 $ you are only mixing consonants among themselves and vowels among themselves , so not mixing vowels and consonants altogether. So what happened in your solution is that all consonants remained as a single group and vowels remained as a single group. thats the mistake .

Mixing all 7 items together $\neq$ mixing within 2 groups of 4 items and 3 items among themselves

$$ 7! \neq 4! \times 3!$$

Idea is

first choose $7C4 \times 5C3$ ,

then arrange the things that are chosen, so as we chose 7 items , arrange all 7 in $7!$ ways . so you will get

$ choose\ and\ then\ arrange\ $ = $ 7C4 \times 5C3 \times 7!$


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