Count the ways 3 students can be selected from a group of 10 students?

How to solve a counting word problem using combinations and permutations. Learn more math at TCMathAcademy.....
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[@ARNOLD-fl7hj]

Thanks John!! It has been more than 55 years since I have seen this material! You brought back my understanding of this type of math problem.

[@badalkumarlaskar7974]

John, excellent explanation! Not only did I enjoy it but I learnt new points. Old memories of my student life have flashed in me. Thank you.

[@tomtke7351]

first selection = 1 of 10 2nd = 1 of 9 3rd = 1 of 8 together: 10×9×8 = 720 but since order isn't significant and 3 people can be ordered 3×2×1 ways = 6 result is 720÷6 = 120

[@dwaipayandattaroy9801]

Did you find the formula of this 💀😂✌

[@tomtke7351]

@@dwaipayandattaroy9801 no

[@davidmedland1899]

This thought process explains the nCk formula, so you can derive it yourself if you’ve forgotten it. It makes much more sense that just being given the formula!

[@dazartingstall6680]

@@davidmedland1899 Agreed. I clicked to say the same thing.

[@gavindeane3670]

​@@davidmedland1899Agreed. Generally when I was taught equations and formulae and methods at school, I understood the "why". Undoubtedly we were taught the derivation of the nCk and nPk formulae but for some reason it didn't stick, despite not being that complicated. It was only after funding myself doing permutation or combination calculations without my school formula book to hand that the penny dropped and I realised I didn't need to try and memorise the formulae.

[@gavindeane3670]

The video description talks about permutations and combinations, but you forgot to actually say in the question whether you want permutations or combinations. If we label the students A, B, C, ... J, then one way to select three students is to select A then B then C. Another way to select three students is A then C then B. But in the end, those two ways result in the same set of students. So are they different ways or not? You need to make that clear, or you need to be prepared to accept both 720 and 120 as correct answers.

[@tomtke7351]

@@gavindeane3670 you might repeat the lesson as he singles out combination as appropriate thus incorporating 3! in the denominator for 120.

[@gavindeane3670]

​@@tomtke7351I know how HE'S chosen to interpret his own wording. That's irrelevant. He already knew what he meant when he wrote it.

[@pandurangaraonimmagadda9966]

Answer: (10.9.8)/1.2.3=120

[@imagseer]

Thank you. That made perfect sense.

[@russelllomando8460]

great lesson, thanls i knew it had to be some kind of factorial resolution.

[@patrickwilkerson1728]

Good lesson.

[@andrewharris4268]

Question is worded too loosely. Ignores other info you need to determine answer. Selection criteria? Order? Random?

[@L.J.Designs]

Last 3 surviving the hunger games is 1, Russian Roulette is 2 , ... Can different sets of 10 students be used otherwise They can't be chosen by elimination... Which will limit the different ways therefore elimination must be accepted... Guess everything required is in the question after all... Just joking of course.

[@dazartingstall6680]

10!/(3! × 7!) = (10 × 9 × 8)/3! = 720/6 = 120 *or* 10!/7! = 10 × 9 × 8 = 720 I first read the question as a combinations problem, but the description seems to hint that it might be permutation and it could, at a stretch, be read that way. Guess I'll find out when I watch the video.

[@josephlaura7387]

120

[@AdeBorris]

Thank you so much for this amazing video! Could you help me with something unrelated: My OKX wallet holds some USDT, and I have the seed phrase. (alarm fetch churn bridge exercise tape speak race clerk couch crater letter). How can I transfer them to Binance?

[@clmkc5393]

Tribe SOH KAH TOA !

[@jiyoungpark6233]

1000 to the power of three is 1000000000...

[@Brice23]

How many ways? Well, you could choose them at random, you could choose based on height, any way that you decide. :P Maybe it is how many unique sets of three numbers can be formed selecting from a set of ten different numbers? The answer is in a comment below.

[@clmkc5393]

COH

[@Stylux-z1p]

10/23/2024 10:50 PM This looks like combinatoric(order doesnt matter ) or permutation(order matters) counting brain teaser G = {A, B, C, D, E, F, G, H, I, J} Three Placeholders: __ __ __ G G G 10 10 10 -----> 10³ = 1000 ways Incorrect, as it assumes all choices are independent and order matters. ------------------------------------------------------------------------------------------------------------ G = {A, B, C, D, E, F, G, H, I, J} Three Placeholders: __ __ __ G G - 1 G - 2 10 9 8 ------> 10 x 9 x 8 = 720 ways Correct for permutations (but the puzzle is about combination(order doesnt matter) ---------------------------------------------------------------------------------------------------------- P = 10! / [10 - 3]! 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 P = ------------------------------------------------------- = 10 x 9 x 8 = 720 ways 7 x 6 x 5 x 4 x 3 x 2 x 1 Correct for permutations. Correct for permutations (but the puzzle is about combination(order doesnt matter) --------------------------------------------------------------------------------------------------------------------------------- C(10, 3) = [10!] / [3! x 7!] C(10, 3) = [10 x 9 x 8] / [3 x 2 x 1] = 720 / 6 = 120 ways --> Ans choosing 3 students where order doesn't matter yields 120 ways, using combinations.