Combinations with restrictions

Simple, easy to understand math videos aimed at High School students. Want more videos? I've mapped hundreds of my videos to the Australian senior curriculum at my website mathsvideosaust...

[@pgt3537]

Biloela Spec class of 2021 appreciate your efforts

[@JoelSperanzaMath]

So happy to hear from you! My best friend when I was a high school student at boarding school was from Biloela.

[@user-zu5ym4uk7z]

your content is a massive help

[@areilavakem4777]

hello, i always confuse combination with permutation, i know their definition , that perm. means there is distinctions in selection, but I still have hard time differentiating them.

[@shlokavasant7471]

You’re a life saver

[@simohayha6160]

Thanks !! u provide difficult questions unlike others!!

[@d6853]

Hi, let’s say I have 4 letter F’s, 4 letter D and 1 H, How can I pick 3 letters and find out how many combinations there are without two (or more) D’s next to each other. I want to treat each letter as a completely separate element so I don’t mind if some combinations are counted more than once because I’m finding permutations. Eg - DFD, FFH etc are accepted but DDF is not, D’s cannot be next to each other. Also, just to complicate it further, The letter F can be next to each other in 2/3’s of all cases. So if F is next to each other in 12 of these cases then you can only count 8 of the ones where F is next to each other. And out of all the ones left you only count half of the ones where a F or a D are next to each other. I’ve been really stuck in this problem!!! You don’t have to answer it for me but could you try to steer me in the right direction? Thanks!

[@aadiyajwas5275]

thank u so much!!!

[@atirahnordin2457]

for question 26, c. at least 1 man, shouldnt it be other alternatives, such as 2 men, 3 men, 4 men and 5 men? like for d, question as for at most 1 men, so, 1 alternative is no man and 2nd alternative is 1 men.

[@JoelSperanzaMath]

There's often more than one way to answer a question. The way I did question 26c relies is a "shortcut", because the compliment of at least one man is no men. If you take all the combinations, and subtract the combination with no men, you get the remaining combinations, which is those with at least one man. Your way would work as well, but requires 4 calculations, whereas mine only requires two.

[@atirahnordin2457]

@@JoelSperanzaMath ah i get it now. Thank you very much👍🏻

[@cark6176]

for 25 it asks as how many teams can be made without sam or tess, shouldnt you take into account the teams made with one or the other. instead of subtracting 6c2, shouldnt it be 7c3 + 7c3 thst you subtract.

[@JoelSperanzaMath]

The question was "how many teams can be made without Sam & Tess". To solve, I've calculated the total number of teams, and subtracted the number of teams with Sam and Tess on them.

[@muhammadadnan0075]

Hi, Thank you for the video. May i ask what book are you taking these examples from?

[@Shan0215]

Cambridge specialist Mathematics y11

[@emperoy8]

Which course is that?