Permutations & Combinations : Help is on the way!

Ian7777, can you pls explain how the answer is 2 x4x3x2x1? Thanks

Yes. There are 5 seats in the car. You need only ask yourself logically how many options there are for each seat. The first seat is the drivers seat, so there are only 2 who can sit there. Once one person sits down, however, the remaining 4 are available for the second seat. Then there are 3, and then 2, and then 1. We just multiply that all together for the answer.

I dont mean to be presumptuous here but there is a formula for that. Yes it needs to be developed from logic definitely. WHat i felt the formula in this case is


out of n r are of one type and n-r are the other type. while r can replace n-r but n-r cannot replace r .

Now it reamins to be seen how mnay places the kind "r" alone needs to occupy. That will help to formulate. Lets call it k ( r>k) . Then the places for n-r objects of one kind and r objects of the other kind can be arranged as

[ n-r -(r-k) ] ! * r!
to test ,n = 5 r =2 k=1
n-r -(r-k) = 4

4! is the number of ways the passengers can be seated. there are 2 drivers , therefore the number of ways is 4!*2!

It can b chekced with similar arrngements . seems to work with circular permutations too.

Ok before i ruffle anyone completely . I did this purely for academic reasons. I know it is lengthy and screwy.. but yeah if i can develop a habit of finding patterns early, i may not have to actually write down the formula , but can bring out the logic really fast. Once again not trying to show off!!!!

ok so say you dont want to use the formulas can u give an example of when something might be a permutation rather than a combination and the logic for figuring that out vs. the logic for figuring out a combination?

yes. spend some time with the concepts of permutations and combinations and learn why one is different from the other. You'll see that combinations are just permutations with an extra division. So go back further and learn why permutations work. They work because when we multiply all the options together, simply, we get the total number of outcomes. So be logical and simply figure out what the options are for each piece of the arrangement, and you'll have the answer. If the order doesn't matter, just divide that answer by the number of spaces factorial. If you're not sure why, learn why. It's because you're canceling out all the repeated arrangements and are just left with the distinct groups.

When you know why these things work, you have power, knowledge, and an edge over this test. When you just learn formulae, you have the potential to make a really big mistake.

yep, probably the formulas are not necessary after all but knowing how to solve perm and comb problems is needed.

I commit the mistake of thinking that I was not going to need it to get 600+; however, on test day the fifth question was about perm and comb

Imagine you order a pizza, and you want a three topping pizza. You will have pepperoni, mushrooms, and onions. The waiter asks you, "which topping do you want on your pizza first?"

What would you answer? Would you ever expect this question? No. Because the order of the toppings wouldn't matter. Your pizza will have three toppings, end of story.

Now, if you did want to know how many ways you could put those three toppings on your pizza, you could figure it out with permutations: 3 x 2 x 1 = 6. But at the end of the day, they're all the same, so you don't count them.

What would be different? Well, a different pizza, that's what: sausage, pinapple, and onion would be a different pizza. Again, you wouldn't care about the order, but it would be different.

That's the difference between permutations and combinations. In permutations, we count all the orderings of every different group. In that case, the two pizza examples would be 12 different permutations. But in combos, we don't care about the order, so we neutralize the copies. Take that 12 permutations from above, divide by the number of copies of each group, and that's combos, in this case: 2.

So combinations is a permutations with a tax on it. You manipulate a permutation question to get a combinations answer. And you do it by dividing by the number of spaces in the permutation, factorial.

And that's how it goes. Look for examples of each on the web - the common divisor between all combos is that we won't care about the order.

Permutation
Definition: An arrangement of objects whose order is important.
Example: Permutation of objects A B C results
ABC, ACB, BAC, ACA, CAB, CBA
There are 6 orders, or 6 permutations in total

Factorial:
Example: The total permutations on ABC is
3!=3 X 2 X 1=6

Example 2: To find all permutations on A B C C
Answer: ABCC, ACBC, ACCB,
BACC, BCAC, BCCA,
CABC, CBAC, CBCA,
CACB, CCAB, CCBA

There are 12 orders, or 12 permutations in total.

P( permutation of n objects with n1 identical ones)
= n!/ n1!
Example 2 of ABCC:
n = 4; (A,B,C,C)
n1= 2; (C,C)
n!/ n1! = 4! / 2! = (4 x 3 x 2 x 1)/ (2 x 1) = 12

Here is my list to help P&C and probabilities:


Start Here-
Overview of Combinations: https://regentsprep.org/regents/math/math-topic.cfm?TopicCode=combin
Overview of Permutations: https://regentsprep.org/regents/math/math-topic.cfm?TopicCode=permut

Advanced-
More Permutations: https://www.themathpage.com/aPreCalc/permutations-combinations.htm#perm%20
More Combinations: https://www.themathpage.com/aPreCalc/permutations-combinations-2.htm

Came across this question on this site itself and I am struggling to get the answer the way they describe it and why my approach is at fault

here is the question:

EXAMPLE 10 A three-person committee must be chosen from a group of 7 professors and 10 graduate students. If at least one of the people on the committee must be a professor, how many different groups of people could be chosen for the committee?

A. 70
B. 560
C. 630
D. 1,260
E. 1,980


My approach :

Since we need to have a professor for the 3-member committee , we can choose the first person as professor in 7 ways .
Rest two positions can be filled with remaining professors or students in (16 C 2 )

total number of combinations for filling 3 member committe -
first position with professor * rest with 2 from 16 mombers ( 6 professors + 10 students) is
7 * (16 C 2) = 7 * (16 * 15 )/2
= 840

but what is baffling is total number of combinations for choosing 3 members from 17 ( 7 professors + 10 students) without restriction of atleast one professor in the committie = 17 C 2 = 680

how can this combination be less than restricted combination

can someone explain me where my thought process took a wrong turn ..

This is a great question. Your problem is a very common one. I've seen it a lot. I hope this discussion will help a lot of people to clear things up.

In your approach, you first chose 1 professor. Then you chose the rest two people from the rest 16 people. The problem is that you would have some repeats here. For example, you may have chosen P1, and then P2 and S1. But it is also possible that you would have chosen P2, and then P1 and S1. In your approach these two cases are counted as two outcomes. However if you look at them, it's the same group of three people. What you should have done, is to break it into three cases, where 1, 2, or 3 professor(s) would be included, so that you know for sure you will not be double counting.

I am also in dire need of help with this one:


In a certain game, a large container is filled with red, yellow, green, and blue beads worth, respectively, 7, 5, 3, and 2 points each. A number of beads are then removed from the container. If the product of the point values of the removed beads is 147,000, how many red beads were removed?
(A) 5
(B) 4
(C) 3
(D) 2
(E) 0

Answer is D. See below link

https://gmatclub.com/phpbb/viewtopic.php?t=31621

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Came across this question on this site itself and I am struggling to get the answer the way they describe it and why my approach is at fault

here is the question:

EXAMPLE 10 A three-person committee must be chosen from a group of 7 professors and 10 graduate students. If at least one of the people on the committee must be a professor, how many different groups of people could be chosen for the committee?

A. 70
B. 560
C. 630
D. 1,260
E. 1,980

=============================================
BLISSFUL wrote:
Easiest way to solve the above Q is:

Total ways of choosing 3 from 7 profs & 10 students = 17C3
Total ways in which there is NO Prof = Total ways in which ONLY students are there = 10C3.

Now Total ways in which ATLEAST ONE prof is there is:
Total possible ways of choosing 3 from 17 - Total ways of choosing all 3 as students( In all other selections, there will be ATLEAST one prof present )= 17C3-10C3 = 560 Ans B

I arrived at the correct answer but when trying the alternative method I cant understand why it doesn't work:

Number possibilities for 1st member = 7
Number possibilities for 2nd member = 16
Number possibilities for 3rd member = 15

So why not 7*16*15/3*2*1 = 260????????

It's getting late - I need to sleep!

Here is an example to something similar I saw in one of the OG or Kaplan books that I can show the difference between a combination and a permutation. Two versions of the question.

Permutation Version:

There are 11 chemicals jars in a laboratory that are identified by a two color labeling system. How many colors are needed to identify all 11 chemicals?

The answer is: 4 colors
Quick solution....
-4 labels using the same color twice
-11-4= 7 labels remaining that will use two different colors
-4*3 = 12 additional color combinations so this is sufficient

Longer solution...
-You know you will need significantly less then 11 colors to do this so pick a number like 3 to start with.
-three colors(R, G, and B) will give you three labels with: RR, GG, and BB
-You still need 11-3 = 8 labels.
-The resulting possibilities of mixing the three colors are: RG, RB, GR, GB, BR, and BG = 6 labels
-3+8 = 11. Not enough possibilites.
-So, 3 is barely too small. Work up from there. Let's try 4.
-four colors(R, G, B, and Y) will give you four labels with: RR, GG, BB, and YY
-You still need 11-4 = 7 labels.
-The resulting possibilities of mixing the four colors are: RG, RB, RY, GR, GB, GY, BR, BG, BY, YR, YG, and YB = 12 labels
-4+12 = 16 label possibilities which is sufficient.

Combination Version:

There are 11 chemicals jars in a laboratory that are identified by a two color labeling system. How many colors are needed to identify all 11 chemicals if the order on the label doesn't matter?

The answer is: 5 colors
Quick solution....
-5 labels using the same color twice
-11-5= 6 labels remaining that will use two different colors
-(5*4)/2 = 10 additional color combinations.
-10+5 = 15 total possibilities with 5 colors.

Longer solution...
-The same approach applies so we will start out with 4 colors.
-four colors(R, G, B, and Y) will give you four labels with: RR, GG, BB, and YY
-You still need 11-4 = 7 labels remaining
-The resulting possibilities of mixing the four colors are: RG, RB, RY, GR, GB, GY, BR, BG, BY, YR, YG, and YB = 12 labels
-Since the problem said that color order on the label doesn't matter so RG is the same as GR. We have to eliminate the duplicates. 12/2 = 6 possibilities.
-4+6 = 10, which is not sufficient.
-Let's try five colors (R, G, B, Y, and O). These will give you: RR, GG, BB, YY, and OO = 5 labels.
-11-5 =6 labels remaining.
-The resulting possibilities of mixing the four colors are: RG, RB, RY, RO, GR, GB, GY, GO, BR, BG, BY, BO, YR, YG, YB, YO, OR, OG, OB, and OY = 20 possibilities.
-Again, since the problem said that color order on the label doesn't matter we have to eliminate the duplicates. 20/2 = 10.
-10+5 = 15 total label possibilites from 5 colors which is sufficient.


Another example:

Permutation Version:

There are 4 teams in a league. How many different games are possible?

4*3 = 12 games possible.

12, 13, 14, 21, 23, 24, 31, 32, 34, 41, 42, 43

Combination Version:

There are 4 teams in a league. How many different games are possible if two teams only play each other once?

(4*2)/2 = 6 games possible.

12, 13, 14, 23, 24, 34

For combinations you are just getting rid of duplicates.

Hope this helps.