Easy Permutations and Combinations

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801

This just saved me. Your explanations are so soooooooo helpful. I have a quiz tomorrow and I am a lot more confident now. THANK YOU!!!

802

Very halpful. why thank-you for this. i was struggling till i found this.

803

Excellent lectures by which everybody can understand.

804

Hello to the guy who loves math! Can you help me with these problems? Its some examples of permutations…

Three Boys & two girls are to be photographed. In how many ways can they be arranged in a row if,
a. if they are arranged alternatively
b.if no boy sits next to girl
c. if the row begins and end with a boy

805

im confused can you show me how many permutations are in 3 letters and 3 numbers??

806

hi can you share some application problems about these including the fundamental technique and compound probability problems, that would really help :slight_smile: oh and with the answers at the bottom or something like that thanks :slight_smile:

807

I have 5 t-shirt design concepts I want to test (T-shirts A, B, C, D, E) but only want to show a random three to each person in sequential monadic format. I know how many combinations of three can be shown: (5 choose 3) = 10 But if I want to know how many times T-shirt B is in the 10 combinations, what is the mathematical way to do that. I know the answer is 6, but want to be able to figure out when I have more concepts. Thanks you for your help!

808

Hello ! i appreciate your work . Kindly answer and explain this :
You have 200 cards out of which 100 are Male, 100 are Female . What is probability of 2nd Female card before 3rd Male card ?

809

Pretty impressive posts, thumbs up for the great work.

810

@Ian: No problem – break the problem down. Try writing it out, meats are Pepperoni, Sausage, Hamburger. Veggies are Onions, Mushrooms, Peppers, Cucumbers, Olives.

  1. How many pizzas can you order with just one meat topping? (Ignore veggies for now). Someone says “I want a pizza with exactly one meat topping”. How many choices do they have?

  2. How many pizzas can you order with just one meat topping and Onions or Mushrooms? It should be double of 1), since you have the Onion-version and Mushroom-version of each meat topping.

  3. How many pizzas can you order with one meat topping, and any of Onions, Mushrooms, Peppers, Cucumbers, Olives? You should be able to write out the variations.

Hope this helps.

811

permutation and combination is quite confused i always forgt that

812

i should combination and permutation examples

813

Thanks for the help, was having a hard time understanding this concept.
Bookmarked this site for future reference.

814

thanks a lot yar. great

815

OMG! I actually learned it! Thank you so much!

816

Hi Kalid,
Its me Tracey again with another question on combinations. Jaime is the Chairman of a committee. In how many ways can a committee of 5 be chosen from 10 people, given that Jaimie must be one of them?

would I do C 10!
4!(10-4) or do I keep the denominator as 5?

817

thanks a lot!! helped me understand the subject better…

818

OH < THHHHHHHHHHHHHHHHHAAAAAAAAAAAAAAAAAAAAAAANNNNNNNNNNKKKKKKKKKKKKKKKK UUUUUUUUUUUUUUUUUUUUUU SSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSOOOOOOO MUCH. It Was SO FANTASTIC EXPLANATION

819

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 :slight_smile: . Email me if you have trouble understanding my explanation.

820

Hello…I see you solve everyones problems, maybe you can help me out. I need to pick 8 winners of 8 football games…but not just who wins the game, they have to cover the spread. How many combinations are there? thanks!