20. Factorials for ordering. Additional Maths OCR FSMQ

[@alexandrebonnet176]

Hey, I'm currently taking add maths and I came across your blessing of channel through some friends. I really appreciate the videos you have made as I prefer learning on KZbin and working at my own pace. Thank you for making these!

[@bareaumaths]

My absolute pleasure. Hope they help and that Add maths persuades you to take maths (and maybe further maths) at A level. All the best.

[@PianoCubeYT]

Combinations and Permutations are the hardest topic by far in my opinion. Thank you for all the help on it (and all the other material you've provided in the run up to our exams)!!

[@bareaumaths]

Glad it was helpful! Good luck tomorrow!

[@cultofsneks2563]

you my freind are a Godsend, my freind reccomended you to me and so far it's been amazing. We're all so lucky you've done all this for us.

[@bareaumaths]

Really glad the videos are helping. All the best for Friday and beyond!

[@cultofsneks2563]

@@bareaumaths thank you, I’m planning to do further maths so I really want a good grade in ADMaths and these videos have really helped remind me of the content

[@bareaumaths]

@@cultofsneks2563 great - add maths is perfect preparation for further maths.

[@bareaumaths]

I've spotted a mistake in my workings for the question asking for the number of 5 digit numbers when choosing from the 8 cards from zero to seven. Comment here if you know the mistake I made.

[@masterofreality6434]

The 5 digit number can't start with 0 so you would have 7 options for the first slot and then 7 for the second slot

[@bareaumaths]

@@masterofreality6434 absolutely correct!

[@rayyanchan9593]

Hello, Thanks for the videos, I might be wrong but is the mistake you can't start with 0, so it would be 5880 not 6720 for the card question

[@bareaumaths]

Well done!

[@sweetener9286]

@@bareaumaths hi, what is the calculation you do in order to get 5880? thank you (:

[@bareaumaths]

@@sweetener9286 Instead of 8x7x6x5x4, we should do 7x7x6x5x4, because there are only seven options for the first number (as we cannot start with a zero as this would not give a 5 digit number), then once that's filled one has been chosen so there are 7 left for the next number (because now we are allowed a zero), and so on. Hope that makes sense.

[@shadowdragon1892]

@@bareaumaths Are you able to do that with nPr to get 5880 instead of 6720? I am guessing it is not possible?

[@bareaumaths]

@@shadowdragon1892 You could do 7 x 7P4 to get 5880, but sometimes it is easier to think of spaces to fill and how many ways you can do so rather than using nPr.

[@qwerty-ue5zg]

Sir I think you made a mistake at 11:12, wouldn't it be 7x7x6x5x4 (or 5880) since if you put zero as the starting digit the number itself won't be 5 digit?

[@bareaumaths]

@@qwerty-ue5zg absolutely correct!

[@someone...000]

Hey sir could u help me with question 8, part ii), for exercise 10.2, on page 131 pls. I keep getting 546 but it says it's 1296, and i really dont understand how. My logic is that if each of the 6 compartments can hold at least 4 toys, and we are allocating the 4 toys into the 6 compartments, then the ways they can be arranged are: we allocate 4 toys all into one compartment (6 ways of doing so), and if we allocate 3 toys into one compartment, then there are 6 ways of doing this, and then we have 1 toy left, and so we have 5 compartments left, thus 5x6=30. And so on for 2 in one, and only maximum 1 toy in each compartment. At the end my calculation is 6+(6*5)+(6*5*4)+(6*5)+(6*5*4*3)=546

[@bareaumaths]

Ha - I did the question just now without looking at your thoughts, and got 546 as well. There are two possible options - either the book is wrong (no textbook I know is 100% accurate in the solutions pages), or we are. Either way, I wouldn't worry right now, and simply be happy that your logical thinking makes excellent sense and the exam question on this topic (if there is one) is not likely to be this difficult. If anyone else can see where they got 1296 from - I'd be glad to hear it!

[@hunnyisstudying]

Hello sir, could you please tell me how to do question 9, exercise 10.1, page 126? I've tried everything but I just can't get the correct answer.

[@bareaumaths]

Draw a venn diagram. The three sections for A, B (just A, the overlap, and just B) add to 0.8. A in it's entirety adds to 0.5, so you can work out the part of B that isn't in the intersection. Call the overlap X and you know that is 1/4 of the entirety of B, so 3/4 of the entirety of B is the part you've already found. Hence work out x. Then you can deduce the part of A not in the overlap, which is the part wanted for the final answer. It's much easier to explain in person!

[@hunnyisstudying]

@@bareaumaths It's making sense right now (using a lot of brain power). I will try it. Thank you!

[@F.Sarpong]

How do you get the answer to question 3)ii) and question 5) in exercise 10.2

[@bareaumaths]

If you can use as many 9s, 5s, 1s, 4s, 2s, 3s, 8s, as you want, then you have 7 choices for the first digit, 7 again for the second digit, 7 again for the third, etc..., which gives you 7^6 or 117649 possible numbers you can create. For question 5, you need 3 digit numbers, so they start with any of 1,2,3,4,5,6,7,8,9, then the second digit can be 0,1,2,3,4,5,67,8,9 and then the third digit can be either 0 or 5. So you do 9 x 10 x 2 - which gives 180 possible options. Does that help?

[@H1289-h8o]

Hi I’m really struggling with the exercises on 10.2 despite watching your video and others, do you think I should move onto permutations and combinations then go back to this? Edit: the ones I am struggling with are from question 6 onwards which are problem solving.

[@bareaumaths]

Yes - watch the videos on permutations and combinations - they should help. This is one of the toughest parts of the course. The questions can get really different and they don't fall into a predictable pattern. It takes a lot of experience to be able to master this topic. I know a lot of maths teachers who struggle with this chapter. Don't lose heart!

[@H1289-h8o]

@@bareaumaths Thankyou