How many ways can 3 students be chosen from a group of 10 students?

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How to solve a combination math word problem. Learn more math at TCMathAcademy.com/.
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26 мар 2024

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Комментарии : 96

@scottekoontz 3 месяца назад

As a former math teacher I know that the question was meant to be a combination, but linguistically the way it is worded the answer to "how many *ways* can 3 be selected" it is a permutation. Selecting A then B then C is different than selecting B, C, A which is a different way. Not sure how to remedy the situation but the questions are sometimes difficult for students who read it literally.

@andrewcox92867 3 месяца назад

How many teams of 3 students can be selected out of 10 students ?

@adrianm.2043 3 месяца назад

It's nearer sixty years than fifty years since I did this at school, but when I read the question I thought exactly the same as you, only thing is I didn't remember it was called a permutation.

@terrycameron9728 4 месяца назад

I love your lessons!

@bulldog6925 4 месяца назад

To help think of answer, think this way: first- there are no restrictions on selection process. So, first person has 1 in 10 chance of being selected because there are 10 students. Second person selected has a 1 in 9 chance of being selected because there are 9 students. The third person selected has a 1 in 8 chance of being selected since there are 8 students left. It does not matter WHO is selected each time, just that with each selection the number of possible choices is reduced by one. To determine the ways that any three students can be selected (whomever they are) is found by multiplying the 3 selection processes ; i.e., 10*9*8 = 720 ways. This is part of set theory - if interested, look up the difference between combinations (which the problem asks) versus permutations (which the author explains)

@shanejohns7901 3 месяца назад

I am taken back to urns and colored balls (both with and without replacement), but I do believe you are correct. I had my university 'finite math' class in the early 90s. The question is misleading, at best. Why should we agree that the CHOICE of A,B,C is identical to choices A,C,B, C,A,B, C,B,A, B,A,C, and B,C,A? In fact, in any grouping of 3 students, there will be 6 different groupings (3*2*1). But if he insists that's the same group, then the answer will need to be 720 divided by 6, which gets us to the 120.

@thomasmaughan4798 3 месяца назад

That was the question that popped into my mind; permutations versus combinations.

@andrewcalvert3875 3 месяца назад

There are lots of ways you can select three students from a group of ten students. You could pick the three tallest. It the three lightest. Or the three youngest. Or the three who did the best of the last maths test!

@scottekoontz 3 месяца назад

This is true. The wording of the question is important.

@MrSeezero 4 месяца назад

10!/3!/7! = 120. If, however, you considered also the order that those same 3 students are selected, then it would be 10!/7! = 720. If order did not matter, then ABC = ACB = BCA = BAC = CAB = CBA. If order did matter, then those 6 would each be different. So, for this example, the multiple of difference between those 2 situations would be 6 or 3!.

@MrMousley 4 месяца назад

OK so your first choice is from 10 and then 9 and then 8 which is 10 x 9 x 8 = 720 meaning that there are 720 combinations BUT in this example A B C and B A C and C B A .. etc .. are the same 3 students so we need to divide by 3 x 2 x 1 720/6 = 120 DIFFERENT combinations.

@bhut1571 3 месяца назад

Yes, that's how my noodle thunk it. And it's an old noodle that used this method in grade 13 (Ontario) over 60 years ago. We were, of course, taught the derivations of the formulae for perms and coms.

@thomasmaughan4798 3 месяца назад

Or (10!) / (3!) (factorials)

@abeonthehill166 4 месяца назад

Great puzzle for the Evening ! Thanks for sharing !

@dougroberts2722 4 месяца назад

Always Great math puzzles!!

@MichaelFenley 3 месяца назад

Ty. Very obscure, but handy formula. Your channel is very helpful.

@philipkudrna5643 4 месяца назад

Very well explained!

@russelllomando8460 4 месяца назад

fantastic explaination. thanks for the fun.

@richardhole8429 16 дней назад

Many times I can do these problems in my head or by joting down a few numbers. This time I have forgotten the combinations formula. So I watched the whole lesson.

@slottibarfast5402 3 месяца назад

You could also consider ways to mean methods like alphabetically, by age, by sex, who is good at math. Randomly, who is the fastest runner, etc. an infinite number of ways to select three out if ten.

@bulldog6925 4 месяца назад

For @charlesmradar - I think I see what you are getting at - sorry if I am wrong. If 1 of the 10 is John, is there a situation where John could be picked three times and the answer is yes. Consider I have a class of ten students and I randomly give out a candy bar to one student once a day for 3 days. Yes, the possibility exists that John could be picked for each of the three candy bars in which case the chances of that happening would be 10*10*10 or 1 in a thousand. The resulting set, though, of who got a candy bar would just consist of one member, that being John.

@dr_ned_flanders 3 месяца назад

I got this in a much simpler way. There are ten students you can choose for the first, nine for the second and eight for the third. This is 720 but there are six ways of arranging three students so divide by 6 and you get 120.

@samueljohnson5947 3 месяца назад

120 is the answer!!

@marvinzorn9907 4 месяца назад

21:29 he says "10x9x8, times 3 factorial". Shouldn't that have been "divided by 3 factorial" ?

@pas6295 4 месяца назад

Correct. Not to occur repeat.

@stompthedragon4010 4 месяца назад

I'm feeling like I hate math more then before, but since I also really want to understand, and I like the way you go thru the problems I'm going to stick with it for awhile.

@tomtke7351 3 месяца назад

And unlike going to a live class, here you can backup and repeat as often as needed and no one else is impacted. Also, John is guilty (in a good way) of saying too much rather than too little... This helps to communicate better to support better learning. Also know you can speed up playback if you'd like.

@seibertmccormick184 3 месяца назад

I got 120. I knew it immediately because all of the other numbers were way too low. let's call the students a, b, c, d, e f, g, h, i and j. The first combination is abc.then abd, abe, There are 8 ways with the first students being ab. There are 7 ways for ac. There are 6 ways with ad. Remember, we are only counting new combos. There are 36 combos where the first student is a. Then we go through the first student is b. There are 28 combos. Then count the ones where the first student is d. There are 21 more combos. Then for e, there are 15 more. for f, there are 10 more, etc. Add them all up and it comes to 120.

@zewdugebre-hiwet2318 3 месяца назад

I don't know the formula but I used the same logic you used and came up with 36 combinations, i.e. 8+7+6+5+4+3+2+1=36. 120 is wrong.

@bulldog6925 4 месяца назад

The question asks how many ways can 3 students be selected ... which is 10X9X8 = 720

@RobNMelbourne 4 месяца назад

Agree, I had a problem with that too. Semantics is important. It would be grounds for a protest if an exam question was worded that way. (I have an MSc in Applied Stats)

@charlesmrader 4 месяца назад

Take that one step further. You could select John all three times, etc. "Select" is not the same as "Do a selection".

@johnplong3644 4 месяца назад

@@RobNMelbourneyeah I had 2 classes in Statistics It has been awhile I was like why am I coming up with a different answer one that is not even one of the solutions.

@bulldog6925 4 месяца назад

@@charlesmraderNo. When John is selected, he is taken out of the selection pool and placed in the set pool. So he can only be selected once. That is why when a selection is made, that person is removed from the selection pool and placed in the "result" pool.

@williamnachtrab9787 4 месяца назад

No, the answer is 120 because in a selection problem order doesn't matter. If the question was how many ways to arrange 3 students drawn from a group of 10 then it is permutation problem because order matters. It's the difference between arrange and select. If you play poker, how many five card hands are there in a deck of 52 cards? Ans: 2.599e6. Because the order of the cards in your hand doesn't matter. Dice problems on the other hand are permutation problems because order matter. eg. 6 and 1 is different from 1 and 6.

@merlenestewart2147 4 месяца назад

I wish the audio was better. I unable hear it's so low.

@warrenporter2331 3 месяца назад

I recall you giving the definition of factorial for any number (10! = 10 * 9 * 8 * .... * 1) but not WHY it works. Something like "If we take, say five unique items, how many ways can we put them in five vacant positions? The first item can be placed in any of the 5 vacant spots, the second item only has 4 spots to choose from, the 3rd item can go into only 3 slots, there will only be 2 slots for the next to last item to fit in, and the last item can only go to the slot that is left. To determine the number of ways to arrange these five items, we would multiply 5 (for the first item) times 4 for the second, times 3 times 2 and then times 1. This is called five factorial etc. " I think that might make what comes later easier to grasp.

@Serisky621 3 месяца назад

D) 120

@KW-gb9cd 3 месяца назад

How many ways can they be selected? Well, there's eenie-meenie-minie-moe, for one . . .

@tomtke7351 4 месяца назад

does order matter? as in .. are the following considered same or different #4 #8 #2 #2 #8 #4 #8 #2 #4 ???????????? otherwise, first choice is 1 of 10 second choice 1 of 9 third is 1 of 8 10×9×8 = 720 if order is unimportant 720/3! =720/6 =120

@srr9281 3 месяца назад

The best reply so far to ‘parse’ the question. As mentioned further above, as basis to challenge the question, not the answer that results.

@pollyj4jam157 3 месяца назад

Yes I pictured the students so I appreciate the lesson.

@Life123love1 3 месяца назад

I’d like to see the derivation of the formulas

@zewdugebre-hiwet2318 3 месяца назад

If you take any two students out of the 10 with a different student, you come up with only 36 combinations. AB will have 8 combinatios, BC 7, CD 6, DE 5, EF 4, FG 3, GH 2, HI 1 which add up to 36.

@oahuhawaii2141 3 месяца назад

If the order of the students does matter, then the answer is 10!/(10-3)! = 10*9*8 = 720. This isn't in the choices for answers. If the order of the students doesn't matter, then the answer is 10!/(10-3)!/3! = 10/1*9/2*8/3 = 120. This is choice "d".

@pandurangaraonimmagadda9966 4 месяца назад

10(c,3)= 10.9.8/1.2.3= 120

@cbargainer 3 месяца назад

There are 720 outcomes; but there are an infinite number of ways to choose them. Height, weight, age, IQ, location, orientation, blood type, etc. Word problems are hard because they are almost never specific enough. Students have to learn what assumptions are accepted by teachers, and that isn't what the students think the lesson is about.

@pulsar22 3 месяца назад

While the REASON for selecting may be infinite, the set of 3-student groups is finite and an exact number. So, whatever reason for selecting that 3 student group out of the 10 students, the number of combination can be calculated.

@bartconnolly6104 3 месяца назад

Surely to create a group of three there are 10 options for the first then nine for the second , as you already picked one from the original,ten then eight to chose from for the third which is 10x9x8 =720 ? But that option isn't in the answers!

@bartconnolly6104 3 месяца назад

Oh no,that's permutations not combinations. I'm recalling something likeNC R and N PR?? The PERMUTATIONS are 720 but you divide this by (N-R) ! OR (10-7)! =3! =3×2×1=6 720/6 = 120 combinations.

@JMM440xi 4 месяца назад

Found this problem interesting.

@stephenremo9200 3 месяца назад

Anyone who bets at horse racing knows that it is 10x9x8. 720 if order matters as the question was asked

@biscotty6669 4 месяца назад

I'm not sure how more clearly the question could have been put. How many unique groups of three people could be chosen? Order doesn't matter and people can't be chosen twice (which would violate a whole bunch of physics). Anyway this is just "n choose k". The math seems surprising at first, perhaps. Check out the binomial theorem on Wikipedia for a thorough explanation. Most scientific calculators even have a button 'ncr' with a capitaized c to calculate this.

@aryusure1943 4 месяца назад

No guess here because it's a rerun. ;) I remember the answer. But I need to watch it again because, well, I don't remember what to do. Too many things to remember.

@jonnamechange6854 4 месяца назад

Although it was the same group called by the teacher, it was called in a different way. Sorry but this is ambiguous, so perms not combis

@jarompack7039 4 месяца назад

The multiple choice answers eliminate the ambiguity

@martinbennett2228 4 месяца назад

This is unnecessarily complicated. The first selection is 1 from 10, the second 1 from 9 and the third from 8. This gives 720 (10 x 9 x 8) possibilities. For the three students there possible arrangements is 1 x 2 x 3, (ABC, ACB, BAC, BCA, CAB, CBA) so each group appears 6 times. The answer is 720/6 = 120.

@paulhammond6978 3 месяца назад

If you are going for the general formula, the most elegant way to specify is to say it that way, otherwise you get into - right, you want to do n. (n-1). (n-2). ... (n- k + 1), which involved assuming that you see you take one away from n until you get to n-k, and multiply them all. If you say the number you want is n! / (n-k)! you have specified that without having to use the ellipsis and assume the person sees what the pattern is to get the result.

@theironherder 3 месяца назад

I guessed the correct answer because it was the only factorial result.

@danielmadden9691 4 месяца назад

d)120

@CarmenRosas-ds5rn 3 месяца назад

@mikebolt3753 2 месяца назад

10! divided by 7!

@bhumibhumi1470 3 месяца назад

120

@pas6295 4 месяца назад

d is the answer

@pas6295 4 месяца назад

120 ways

@larsnystrom6698 3 месяца назад

The correct answer is 720. But then you think the 6 possible permutations of those three students matter, so you think it's 120 ways. That is of course wrong!

@peta1001 3 месяца назад

Look at the moment on the board at 12:38. I cannot imagine a mathematician who believes this is the definition of Permutations and Combinations. The only difference is P and C...according to you. I thought a mathematician should be detail-oriented... Who knows?!?

@robertkelley3437 4 месяца назад

You are a Pizza shop. You are making pizzas. A customer come in and order one each of every possible combination of pizza. There are 8 toppings A, B, C. D, E, F, G, H plus just a plain bread pizza. How many pizzas will you have to make using any toppings from 1 to 8 of the toppings? What is the formula?

@dwaipayandattaroy9801 4 месяца назад

Its 10!/ (10-3)! = 120

@herbertklumpp2969 4 месяца назад

10 over. 3

@thomasschodt7691 3 месяца назад

(10)! / (10-3)! / (3)!

@raynewport9395 4 месяца назад

Another attempt at the English language goes awry.

@beverly987654 4 месяца назад

Once again the math people need to go back to an English and communication class. There are infinite ways 3 student can be selected from a group of 10 students. You could select them by age, race, religion, IQ, education level, height etc. etc. etc

@bartconnolly6104 3 месяца назад

But there are not infinite races theremis only one, the human race.

@beverly987654 3 месяца назад

@@bartconnolly6104 Best answer ever.

@nancyholmquist2690 4 месяца назад

It does not make sense!

@bone0944 4 месяца назад

You repeat the same information far too often in an attempt to make the solution clear. Un fortunately it just makes it confusing and unclear.

@pamspencer5377 4 месяца назад

@homayounshirazi9550 3 месяца назад

The language of mathematics is Precise ! ...and you are no mathematician! "senator, I knew JFK" (and you senator are not a JFK.)

@user-qe8gw2xw1p 3 месяца назад

You’re explanation was too long

@stevenk6638 4 месяца назад

your videos move WAY too slow !

@twolery1514 3 месяца назад

What is a "way"? To explain what I mean, I will offer three "ways": (1) random, (2) by weight, and (3) by height. Sir, your math may be great, but your use of the English language is not. Please think of how to explain the problem you have in mind in a more precise "way" so that other possible interpretations are eliminated.