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GMAT CH 21: Combinatorics/Counting Method
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Terms in this set (37)
Therefore, the total number of arrangements of n different objects in a row is
N=n∗(n−1)∗(n−2)....2∗1=n!
COMBINATON: A combination is an unordered collection of k objects taken from a set of n distinct objects. The number of ways how we can choose k objects out of n distinct objects is denoted as:
A permutation is an ordered collection of k objects taken from a set of n distinct objects. The number of ways how we can choose k objects out of n distinct objects is denoted as:
P/k! = n!
Circular arrangements:
Let's say we have 6 distinct objects, how many relatively different arrangements do we have if those objects should be placed in a circle.
The difference between placement in a row and that in a circle is following: if we shift all object by one position, we will get different arrangement in a row but the same relative arrangement in a circle. So, for the number of circular arrangements of n objects we have:
R = (n-1) !
TIP: USE P OR C FORMULA?
NETIEHR. DO NOT APPROACH THESE PROBLEMS WITH THE MINDSET OF A FORMULAE.
Any problem in Combinatorics is a counting problem. Therefore, a key to solution is a way how to count the number of arrangements. It sounds obvious but a lot of people begin approaching to a problem with thoughts like "Should I apply C- or P- formula here?". Don't fall in this trap: define how you are going to count arrangements first, realize that your way is right and you don't miss something important, and only then use C- or P- formula if you need them.
How to explain P and C in words
Permutations are ordered. nPr is how do we arrange n objects r at a time? There will be multiple unique ways of doing so.
Combinations are not ordered. It does no matter. For example: How many times can you fill 4 boxes with the letter ABCD?
=> First calculate permutations: There will be 6p4 = 360.
In these ABCD and DCBA is part of one combo bc the same 4 letters are there. So we have to remove "the no of times r can be arranged in r places (or R1)" from the permutations. This gives us combinations.
Moral: Do not remmeber either as a formula. Only thing to remmeber is no fo ways n can be arranged is n!. If n has be arranged r at a time we have nPr. When r has to be chosen from a pool of n we have nPr/R!
The Fundamental Counting Principle states that if an event has x possible outcomes and a different independent event has y possible outcomes, then there are xy possible ways the two events could occur together.
The no of ways to arrange 6 (or n) objects is 6! (or n!)
For example, how many three-digit integers have either 6 or 9 in the tens digit and 1 in the units digit?
To solve, we need to find the possible outcomes for each digit (hundreds, tens, and units) and multiply them. Each digit has 10 possible values (0 through 9). The hundreds digit can be any of these except 0 (since a three-digit number cannot begin with 0). The tens digit has only 2 options (6 or 9). The units digit has only 1 possibility (1). Therefore, the total number of possibilities is 9 x 2 x 1 = 18.
Critical illustration of why you should not think in terms of formulae but in terms of pure enumeration logic counting and then MAYBE apply formulae if calc is easier
Example 3: In how many ways can you make a five letter password using the first ten letters of the English alphabet? (You can use only capital letters.)
=> you thout its 10p5 = WRONG. This question does nto spcify the condition that once an alphabet is used it cant be used again - which is sort of what the npr formulae is predicated upon. Bc this is the case, the no of options you have every trial is still 10 instead of being 10 then 9 then 8, etc..
So the answer to this will be 10^5 instead of what an npr might give.
Why is counting in a circular arrangement = (n-1)! instead of n! ( which iseen in the case of linear arrangement)
For the first person, all seats are the same so he can choose in 1 way. He creates a frame of reference and thereafter, every seat is distinct (relative to him). So the rest of the (n-1) people can sit in (n-1) seats in (n-1)! ways (using Basic Counting Principle).
In the case of linear arrangement: say there are 4 people eyes 4 chairs => we can write this as 4p4 bc the first person has a choice of 4 2nd has 3... = 4 3 2 1 (multiplied).
In circular arrangeemnt, since the arrangement is RELATIVE the first person is counted as having only one option and based on his choice the relativity of the remaining are established and result in multiple chocies (and not 1 like in the case of the first person)
In what case will circular and linear yield same # of arrangamnets?
Note: In case nothing is mentioned, in a circular arrangement, two seating arrangements are considered different only when the positions of the people are different relative to each other. If it is given that the seats are distinct (say they are different colored), then the number of arrangements is n! (same as in the case of linear arrangements)
Now think of circular arrangement in terms of non linear arrangement.
Question: In how many different ways (relative to each other) can 8 friends sit around a square table with 2 seats on each side of the table?
=> remeMber: there is no rule that the first person has 1 option = the first persons option's are limited to the patterns of arrangement.
This means that in this Q:
If there is a linear arrangement with constraint, what can you do to make it less complex?
Instead of solving the actual question try and mamipulate the question and break it into algebraic parts. Etc: if Q asks you Arrange 6 people in 6 seats but 2 of them cannot sit next to each other
=> an easier way to solve this is (1) find ptemruations for 6 people seating 6 ways and then subtract from this : (2) Find cases where we TREAT the two people as one unit (so essentially they always sit together)
Trick of symmetry in counting: if you can identify similar cases and what fraction do those cases make up of all the cases you can simplify complex problems
Question 6: 6 people go to a movie and sit next to each other in 6 adjacent seats in the front row of the theatre. If A cannot sit to the right of F, how many different arrangements are possible?
Good example of manipulation of word problems in circular arrangeemnt:
Question 1: A group of 8 friends sit together in a circle. If A refuses to sit beside B unless C sits on the other side of A as well, how many possible seating arrangements are possible?
How to differentiate if the WP is a P or a C ?
Focus on what you have to do. Do you have to just SELECT some friends/toys/candies/candidates etc or do you have to ARRANGE them in distinct seats/among some children/in distinct positions etc too.
if you have to only arrange, it is a permutations problem; if you have to first select and then arrange, it is a combinations and permutations problem.
But if you are not using the formulas (nPr and nCr), you don't have to think in terms of permutations and combinations. Just think in terms of selecting and arranging."
Combinations for units is not the same as combinations for groups
Q: Give a kid 4 chocolates out of 12 is NOT the same as
Q: In how many ways can one divide twelve different chocolate bars equally among four boys?
The second one is like individual combinations problems happening simultaneously
Kid 1 gets options (12
11
10)/3! and then kid gets (9
8
7)/3! and so on till kId 4
=> most importantly and means multiplies so each of these cases are multiplied by each other
EVEN MORE IMPORTANT: DISTINCT GROUPS VS NON DISTINCT GROUPS
When the question says that you need to make n groups/bundles/teams that are not distinct, you need to divide by (n!). If the groups/bundles/teams are distinct then you do not divide by (n!).
SORT OF LIKE THE P VS C BUT ADDITIONAL LAYER ^
THINK OF GROUP PROBLEMS AS?
TWO LAYER PERM OR COMB PROBLEMS
=> ALWAYS DRAW FOR THESE.
theoretical reason why 4c3 = 4c1
Example:
The interesting thing to note here is that selecting 3 groups out of 4 is the same as selecting 1 group out of 4. Why? Because we can think of making the selection in two ways - we can select 3 groups from which we will pick a student each or we can select 1 group from which we will not select a student.
The best foormulae to solve complex counting problems is?
NONE.
1) Understand the findamental counting principle
2) Break problem down into cases and maybe simplify cases if you can
3) Then use "formulas" for counting but even then dont thiink about this in terms of plug an dplay actioon but instead everything is perm and then ask yourself aare we arranging or selecting ?
4) You can make different combinations of select then arrange and then unaraange etc to be most logical
Unfair distributions
This is not the same way aas the previous Qs: here, think of it like you are choosing r from a pool of n however one slot is not limited to insertion of one from the pool OR in simple example like terms: ABCD are different people on 5 chairs => but ABCD can sit on top of each other in one of trh chairs
In na unfair dist question, if arrangement does not matter
Then you can use the identical lines method
The two first issues you should always address are: Does order matter, and is repetition allowed?
The two first issues you should always address are: Does order matter, and is repetition allowed?
THINK OF EVERY QUESTION PURELY IN TERMS OF COMBINAATIONS AND TAKE IT FROM THERE
Note that some of the questions above were permutation questions and some were combination questions, but remember, we don't need to worry about which is which. All we need to think about is how to solve the question, which is usually by starting with nCr and then doing any other required steps. Break the question down — select people and then arrange if required. This will help you get rid of the "permutation or combination" puzzle once and for all.
Whenever you look at any Q of distributing items among people or in boxes, the Q will deal with any one of the 4 CASEs.
LEARN to differentiate these 4 groups and you may be able to make OTHERWISE a complicated Q LOOK rather easy..
THE 4 CASES ARE LISTED BELOW
lets differentiate between similar and dissimilar things-
No of balls = 4 and number of boxes=3
1) All balls and All boxes are similar
1) All balls and All boxes are similar
THE NO OF POSSIBILITIES IN THIS TYPE OF QUESTION = NO OF CASES
As you can see in the image, there are 4 cases so there are 4 possibilities.
lets differentiate between similar and dissimilar things-
No of balls = 4 and number of boxes=3
2) Balls are similar BUT boxes are NOT.Now all groupings will be similar to above but the boxes are dissimilar, so lets find ways
2) Balls are similar BUT boxes are NOT.Now all groupings will be similar to above but the boxes are dissimilar, so lets find ways
If the balls are similar think of it as having only 4c4 ways of placement => there is only one way for each type of positioning bc the balls are the same => the only variations in positions come from the arrangement bit
TRICK TO SOLVE THESE: LINE ARRANGEMENT DRAWING METHOD
(MENTIONED BELOW)
lets differentiate between similar and dissimilar things-
No of balls = 4 and number of boxes=3
3) Ball are dissimilar and Boxes are similar
Here boxes are the same but the balls are dissimilar
3) Ball are dissimilar and Boxes are similar
Here boxes are the same but the balls are dissimilar
In this option think of it as
e repeats dont matter = UNLESS THEY ARE IN DENOMINATIONS MORE THAN OR EQUAL TO 2; Bc having 2 groups of 2 when you are making variations gives rise to different possibilities as each object is different (this part is in the paranthesis of the image).
A better example of this type of question is listed on the next card !
A siimilar example to the previous case: Distinct objects placed in similar groups
Take a look at the second last arrangement - only this one is unaraanged while the others are left 9SIMILAR GROUPS Q - ONLY UNARRANGE IF MORE THAN OR EQUAL TO 2)
lets differentiate between similar and dissimilar things-
No of balls = 4 and number of boxes=3
4) All balls and All boxes are dissimilarMANY Qs belong to this category.After finding the ways in third, we work on the solution further
4) All balls and All boxes are dissimilarMANY Qs belong to this category.After finding the ways in third, we work on the solution further
In this you play the combinations game but re arrange all the combinations to account for arrangement => which gives th emost number of possibillities
straight formula here would be 3 can in any of boxes = 3*3
3
3=81
LINE ARRANGEEMNT DRAWING METOD
(TRICK THAT CAN BE USED IN CASE 2: SIMILAR BALLS DISTINCT OBJECTS)
Draw all the objects in a row and seprate them via lines, but make sure that these division do not start at tail ends (first or last place)
CASE 4 EXCEPTIONS TO THE TRICK:
Be careful in certain distinct objects dinstinct groups questions : not all fo them can be solved by places ^ objects trick
Why are these two questions different? After all, we are distributing 5 different things among 4 children/fingers in both the cases. The difference lies in the fact that when a child gets 2 fruits, the fruits are not arranged but when a finger gets two rings, it gives us 2 different arrangements since the rings can be arranged in 2 ways. You can wear 2 rings on your fingers in 2 different ways (A on top, B at the bottom or B on top and A at the bottom). When you get 2 fruits, there is no arrangement involved. Whether you got fruit A first or fruit B first doesn't matter. At the end of it, you have 2 fruits A and B and that's all that matters.
The total number of ways of dividing n identical items among r persons, each one of whom receives at least one item is
The total number of ways of dividing n identical items among r persons, each one of whom, can receive 0,1,2 or more items is
Unless states, things are not identical - generaly speaking
Unless states, things are not identical - generaly speaking
Too many Qs where you are getting counting method with complex constraints wrong. For detailed solution please read this.
When you aare solving problems that involve forming commitees/words/groups (ordered or not)
1) Counting method is always already arranged
2) Split up the required group into twoo different required groups and multiply
3) If the Q asks for two different elements and final group consisting of those elements is ordered => find no of ways you get the first elements and unarrange then find no of ways for second set of elements and unarrange => THEN ARRANGE ALL OF THEM TOGETHER (AND DISCOUNT REPEATS IF NECESSARY)
NUMBER COUNTING VS COMBO PT 2
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