Easy Permutations and Combinations

Hi Cassie,

Your book is right - the answer for the problem is 70.
Below is my explanation from an intuitive approach.

Understanding the quesiton:

As you’ve tossed the coin 8 times in a row, the sample space for the experiment will have 2^8 (= 256) outcomes. Your task is now to count the total number of outcomes that have equal heads and tails(ie.4 heads and 4 tails only) from the 256 outcomes.

Solution:

To solve this, imagine 4 girls(~Heads), and 4 guys(~Tails) playing a simple game. You’ve 8 rooms in a row, and 1 room can hold 1 person only. You have to find out the total ‘combinations’ you can have with 4 Girls and 4 Guys. This will lead to the final answer.
To simplify it further, forget about the 4 guys, and just think of 4 Girls and 8 rooms (the remaining 4 vacant rooms will later be occupied by guys after girls choose 4 rooms for every combination).

So, 8 rooms and 4 girls leads us to -
876*5 = 1680, which is the permutation. But the order of 4 girls(heads) does not matter for the given problem, hence 1680/(4!) = 70.

Hope I did not confuse you :). Email me if you have trouble understanding my explanation.

Hiya! I would be really glad if you solved this problem.
Q. A ‘hand’ of 5 cards is dealt from an ordinary pack of 52 playing cards. Show that there are nearly 2.6 million distinct hands and that, of these, 575 757 contain no card from the heart suit.
On three successive occasions a card player is dealt a hand containing no heart. What is the probability of this happening? What conclusion might the player justifiably reach?

Please answer my question as soon as possible. Thanks…

this is a lot of help, thank you so much:)
but i still don’t understand what k means

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In your combinations example Im still not understanding what happens to the 7. if your not actually multiplying it by the 3 then whats the point of having it there at all?

Thank u very much. u’ve been a great help!!!

Liked it a lot ,the way you arrived at the formula…Great.

Hey, I noticed that you helped a lot of people with permutation and combination problems. I need your assistance.

I need to figure out this permutation/combination formula. How do I add the rule that objects within a particular set obey a hierarchy (i.e., the already predefined list of objects follows the rule that one object cannot appear below another object when it was originally listed above the object).

To better express my meaning clearly, here’s an example:

Set #1 - James, Jennifer, Jason, Amber

In the permuted/combinated list, James shall always appear above Jason and Amber. Amber shall always appear below Jennifer.

I want to add that rule, but I don’t know how.

Furthermore, I want to define this rule for more than one set. I want an equation that will permute/combinate different rules for different sets, but render all possible patterns based on those rules. Each set has 10 items, so the third set will contain only 10 items.

For simplicity sake, I’ll keep the number of items in the list down to six:

Ex. Set #1 - James, Jennifer, Jason, Amber, Florence, Carrie
Ex. Set #2 - Dustin, Michael, Morgan, Lance, Travis, Patrick

The rules are that names will be selected based on a preset pattern (i.e., names will be selected in a 4-6, 5-5, or 6-4 manner). For the first equation, each set maintains their respective hierarchy, but the hierarchy for set #2 does not apply to the hierarchy of set #1. Basically, 4 names will be randomly selected from set #1 and 6 names will be randomly from set #2, 5 names from both, and then 6 names and then 4 names. Once the formula/program solves the permutations/combinations, I want it to take those results and decipher the third set–in essence, match the results against the given but unknown information of the third set.

The third set will have numbers: 2, 2, 1, 3, 3, 4

Each individual in each set will have a rank designation.

Ex. Set #1 - James 2, Jennifer 3, Jason 4, Amber 1, Florence 1, Carrie 2
Ex. Set #2 - Dustin 1, Michael 3, Morgan 2, Lance 3, Travis 1, Patrick 2

Based on given information from the third set, the formula or program will most likely suggest a possible combination that suits the (2, 2, 1, 3, 3, 4) parameter:

James 2, Morgan 2, Amber 1, Michael 3, Jennifer 3, Jason 4

I’m no mathematician, but here’s how I visualize the equation in Microsoft Excel or something:

[(~set #1 & #2 - 4/6)(~set #1 & #2 - 5/5)(~set #1 & #2 - 6/4)] | (set #3)

** Please note that the selection process requires that both sets have 10 items (e.g., 4 items will be selected from set #1’s list of 10 items, 6 items will be selected from set #2’s list of 10 items). Under the aforementioned precepts, this “equation” represents all possible arrangements for these individuals. Then it matches those results and “deciphers” set #3.

I want to do this in excel or in another program, but I don’t know how to manipulate the permutation/combination formulas to obey new rules and so forth. I need help developing a workable formula.

You’re the man. Very helpful. Better than my (foreign) teacher could explain it!

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Very descriptive article, I loved that a lot.
Will there be a part 2?

my teacher kept trying to explain this thing for over two hours … and everytime i looked to my classmate beside me … we were like wtf with the crazy man !!! and we just kept laughing out of confusion !!
YOU ARE A LIFE SAVER …

wow! its a superb site…

I have a question.
you are supposed to tell how many permutations and combinations can be made, im stuck on this question. (im only in 6th grade)

        three of the letters A, B, C, D, E, F, and G

How come you’re the only site that explains why this formula works? Thank you!

very very nice :slight_smile:
if i can understand; anybody can …
GOD Bless yOu !!!

very very nice :slight_smile:
if i can understand; anybody can …
GOD Bless yOu !!!
:slight_smile:

I can compute permutations and combinations all day long, but understanding how and why they work and, sometimes, when to use one over the other (only all the important parts, right?) can be confusing. The phrases ‘order does not matter’ and ‘order does matter’ in the definitions of combination and permutation confuse me. I think about the definitions differently, but want to make sure I am correct. Permutations=order matters=repetitions are okay and do count as new arrangements Combinations=order does not matter= repetitions are bad and do not count as new arrangements Though I KNOW this is faulty, the phrase ‘order does not matter’ computes in my brain as ‘order is not a factor in determining the number of arrangements’. With this thinking, ABC, BCA, and CBA would be considered 3 different arrangements. This, of course, is not what is meant by ‘order does not matter’ in the definition of combination because, in a combination, repetitions do NOT count as new arrangements. Now that I have explained the craziness of my brain, does my thinking about permutations and combinations in terms of repetitions counting as new arrangements or not counting as new arrangements seem valid? Thanks so much for taking the time to help.

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Hi Tracey,

Very close. In this case, since we know Jaime must be on the committee, we’re really only picking 4 other members from the 9 remaining. So it becomes

9 * 8 * 7 * 6 / 4!

or

9! / (5! * 4!)

We divide 9 * 8 * 7 * 6 by the 4! re-arrangements within the sub-committee we’ve chosen.