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From the following list of five question stems, how many are permutations questions and how many are combinations questions?
I. How many ways can ten students arrange themselves in a row of ten seats?
II. How many groups of students can attend a conference if there are ten students and four passes to attend the conference?
III. How many subcommittees of four are possible from a committee of 12 people?
IV. How many ways can Karl arrange his twelve books on a shelf?
V. How many seating arrangements are possible for six couples sitting around a circular table?
A. 1 permutation, 4 combination
B. 2 permutation, 3 combination
C. 3 permutation, 2 combination
D. 4 permutation, 1 combination
E. All five are permutation questions
Could somone please A.) comment each question on the list with the answer to the question and B.) explain how one easily recognises that the question is a permutation or combination problem.
I. How many ways can ten students arrange themselves in a row of ten seats?
= n arragements of n objects = n! = P. The order matters. Take the case of a 3 books in 3 slots. ABC is different than BAC or CBA. Thus the order is important.II. How many groups of students can attend a conference if there are ten students and four passes to attend the conference?
= C. The order is not important. If I have to select ABCD for the 4 passes, then it does not matter whether I choose these 4 people as BACD or ADCB or BCDA etc.III. How many subcommittees of four are possible from a committee of 12 people?
= C, I dont care if the chosen people are a,b,c,d or a,c,d,b or b,c,d,a. The committee of 4 still remains the same.IV. How many ways can Karl arrange his twelve books on a shelf?
= P, the order matters similar to case (i).V. How many seating arrangements are possible for six couples sitting around a circular table?
= CIRCULAR ARRANGEMENT, depending on whether the order (clockwise vs anticlockwise) is important, you will get 2 different answers. Can not be categorized as either permutation or combination.Basic principles are: \(nCr = \frac {n!}{r!*(n-r)!}\) and . Thus you see nPr = nCr * r! ---->
Permutations specify order of selection while combinations do not cater to the order. In other words, if a selection is highly ORDERED, it is permutation, while if the order is NOT important, it will be combination.Example, lets say I talk about a lock that has the code of 123. Then ONLY 123 will work. 321 or 132 or 231 will NOT work. Thus the order matters. This case is of permutation.
But if I tell you that I need 3 fruit out of a mixture of 5, then I do not care whether you pick a,b,c or you pick b,a,c. You will still end up getting a combination of 3 same fruits. In this case, you have combinations.
Read here for more :
math-combinatorics-87345.htmlVeritas prep's book on combinatorics is also a good study material with good collection of problems.
Additionally, most of GMAT combinatorics problems can be solved via counting principle (a simpler way of calculating possible arrangements without the fanciness of 'factorials' or particular formulae of permutations or combinations!).
For combinatorics questions :
PS :
search.php?search_id=tag&tag_id=52DS:
search.php?search_id=tag&tag_id=31Hope this helps.