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

แชร์
ฝัง
  • เผยแพร่เมื่อ 18 ม.ค. 2025

ความคิดเห็น • 107

  • @scottekoontz
    @scottekoontz 9 หลายเดือนก่อน +6

    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
      @andrewcox92867 9 หลายเดือนก่อน +3

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

    • @adrianm.2043
      @adrianm.2043 9 หลายเดือนก่อน

      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.

  • @andrewcalvert3875
    @andrewcalvert3875 9 หลายเดือนก่อน +2

    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
      @scottekoontz 9 หลายเดือนก่อน

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

  • @johnmarchington3146
    @johnmarchington3146 5 หลายเดือนก่อน

    Thanks for the video, John. Permutations and combinations were always my Achilles Heel in maths when I was at school, more decades ago than I care to remember, so I found your video extremely interesting. I won't explain how I solved the problem but it was a real pencil and paper job to be sure and one I definitely couldn't repeat if I were faced with a really large numbers of students. I did partly remember the formulae you mentioned and I know what factorial numbers are. I'm in my 80s now.

  • @bulldog6925
    @bulldog6925 9 หลายเดือนก่อน +5

    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
      @shanejohns7901 9 หลายเดือนก่อน

      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
      @thomasmaughan4798 9 หลายเดือนก่อน

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

  • @MrSeezero
    @MrSeezero 9 หลายเดือนก่อน +3

    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!.

  • @terrycameron9728
    @terrycameron9728 9 หลายเดือนก่อน +4

    I love your lessons!

  • @MrMousley
    @MrMousley 9 หลายเดือนก่อน +4

    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
      @bhut1571 9 หลายเดือนก่อน

      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
      @thomasmaughan4798 9 หลายเดือนก่อน

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

  • @abeonthehill166
    @abeonthehill166 9 หลายเดือนก่อน +1

    Great puzzle for the Evening ! Thanks for sharing !

  • @MichaelFenley
    @MichaelFenley 9 หลายเดือนก่อน +1

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

  • @philipkudrna5643
    @philipkudrna5643 9 หลายเดือนก่อน +1

    Very well explained!

  • @slottibarfast5402
    @slottibarfast5402 9 หลายเดือนก่อน

    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.

  • @dougroberts2722
    @dougroberts2722 9 หลายเดือนก่อน +1

    Always Great math puzzles!!

  • @warrenporter2331
    @warrenporter2331 9 หลายเดือนก่อน

    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
      @Serisky621 9 หลายเดือนก่อน

      D) 120

  • @BasedCarbonLifeform
    @BasedCarbonLifeform 9 หลายเดือนก่อน +2

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

    • @pas6295
      @pas6295 9 หลายเดือนก่อน

      Correct. Not to occur repeat.

  • @bulldog6925
    @bulldog6925 9 หลายเดือนก่อน

    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.

  • @Life123love1
    @Life123love1 9 หลายเดือนก่อน

    I’d like to see the derivation of the formulas

  • @merlenestewart2147
    @merlenestewart2147 9 หลายเดือนก่อน +1

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

  • @russelllomando8460
    @russelllomando8460 9 หลายเดือนก่อน +1

    fantastic explaination. thanks for the fun.

  • @bulldog6925
    @bulldog6925 9 หลายเดือนก่อน +20

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

    • @RobNMelbourne
      @RobNMelbourne 9 หลายเดือนก่อน +4

      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
      @charlesmrader 9 หลายเดือนก่อน +2

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

    • @johnplong3644
      @johnplong3644 9 หลายเดือนก่อน +1

      @@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
      @bulldog6925 9 หลายเดือนก่อน +3

      @@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
      @williamnachtrab9787 9 หลายเดือนก่อน +2

      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.

  • @cbargainer
    @cbargainer 9 หลายเดือนก่อน +1

    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
      @pulsar22 9 หลายเดือนก่อน

      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.

  • @dr_ned_flanders
    @dr_ned_flanders 9 หลายเดือนก่อน

    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.

  • @tomtke7351
    @tomtke7351 9 หลายเดือนก่อน +2

    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
      @srr9281 9 หลายเดือนก่อน

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

  • @KarlSnyder-jh9ic
    @KarlSnyder-jh9ic 5 หลายเดือนก่อน

    Clever. Useful review

  • @richardhole8429
    @richardhole8429 6 หลายเดือนก่อน

    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.

  • @bartconnolly6104
    @bartconnolly6104 9 หลายเดือนก่อน

    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
      @bartconnolly6104 9 หลายเดือนก่อน +1

      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.

  • @zewdugebre-hiwet2318
    @zewdugebre-hiwet2318 9 หลายเดือนก่อน

    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.

  • @stompthedragon4010
    @stompthedragon4010 9 หลายเดือนก่อน +1

    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
      @tomtke7351 9 หลายเดือนก่อน

      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.

  • @KW-gb9cd
    @KW-gb9cd 9 หลายเดือนก่อน +1

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

  • @thomasharding1838
    @thomasharding1838 3 หลายเดือนก่อน

    @ 21:26 You said, "10 times 9 times 8 times 3 factorial". I know you meant "divided by 3 factorial". OOPS!. But still, "How many ways can 3 students be chosen..." does not define whether you are looking for combinations or permutations.

  • @martinbennett2228
    @martinbennett2228 9 หลายเดือนก่อน

    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
      @paulhammond6978 9 หลายเดือนก่อน

      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.

  • @seibertmccormick184
    @seibertmccormick184 9 หลายเดือนก่อน +1

    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
      @zewdugebre-hiwet2318 9 หลายเดือนก่อน

      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.

  • @josephlaura7387
    @josephlaura7387 3 หลายเดือนก่อน

    Thank you

  • @pollyj4jam157
    @pollyj4jam157 9 หลายเดือนก่อน

    Yes I pictured the students so I appreciate the lesson.

  • @pandurangaraonimmagadda9966
    @pandurangaraonimmagadda9966 9 หลายเดือนก่อน +1

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

  • @biscotty6669
    @biscotty6669 9 หลายเดือนก่อน

    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.

    • @gavindeane3670
      @gavindeane3670 5 หลายเดือนก่อน

      The question could have been more clear in that it doesn't say whether order matters or not. It's not clear whether it's asking for permutations or combinations.
      I recognised it as a combinations question but only because I'm used to wording like this being used for combinations. That doesn't mean the wording is clear, it just means that I hated to be familiar with it.
      That's great for me, but it's no use to someone who happens not to be familiar with it.

  • @oahuhawaii2141
    @oahuhawaii2141 9 หลายเดือนก่อน

    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".

  • @jonnamechange6854
    @jonnamechange6854 9 หลายเดือนก่อน +1

    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
      @jarompack7039 9 หลายเดือนก่อน

      The multiple choice answers eliminate the ambiguity

  • @samueljohnson5947
    @samueljohnson5947 9 หลายเดือนก่อน +1

    120 is the answer!!

  • @mouradbelkas598
    @mouradbelkas598 5 หลายเดือนก่อน

    Thank you. Why don't you go directly to the solution instead of many explanations which are outside the problem. Often times, it gets people very confused. This happens with many of your videos. I hope that's not how your teach in class.

  • @johnmarchington3146
    @johnmarchington3146 5 หลายเดือนก่อน

    I spent some time on this one but ended up with the answer (d)

  • @danielmadden9691
    @danielmadden9691 9 หลายเดือนก่อน

    d)120

  • @peta1001
    @peta1001 9 หลายเดือนก่อน

    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?!?

    • @gavindeane3670
      @gavindeane3670 5 หลายเดือนก่อน

      I don't think he's trying to define the terms "permutation" and "combination" there. I think he's just explaining what the nPk and nCk notation means.

  • @stephenremo9200
    @stephenremo9200 9 หลายเดือนก่อน

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

  • @JMM440xi
    @JMM440xi 9 หลายเดือนก่อน

    Found this problem interesting.

  • @aryusure1943
    @aryusure1943 9 หลายเดือนก่อน

    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.

  • @theironherder
    @theironherder 9 หลายเดือนก่อน

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

  • @raynewport9395
    @raynewport9395 9 หลายเดือนก่อน +2

    Another attempt at the English language goes awry.

  • @beverly987654
    @beverly987654 9 หลายเดือนก่อน +1

    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
      @bartconnolly6104 9 หลายเดือนก่อน +1

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

    • @beverly987654
      @beverly987654 9 หลายเดือนก่อน

      @@bartconnolly6104 Best answer ever.

  • @mikebolt3753
    @mikebolt3753 8 หลายเดือนก่อน

    10! divided by 7!

  • @larsnystrom6698
    @larsnystrom6698 9 หลายเดือนก่อน

    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!

    • @gavindeane3670
      @gavindeane3670 5 หลายเดือนก่อน

      He's answering it as a combinations question rather than a permutations question. I recognised it as a combinations question, but that's only because I'm used to combinations questions being worded this way. He would do better to write his question clearly rather than just hoping his audience knows what he meant.

  • @thomasharding1838
    @thomasharding1838 3 หลายเดือนก่อน

    NOPE!! Those three students can be selected only 6 ways. First: student 1, student 2, and student 3. Second: student 1, student 3, and student 2. Third: student 2, student 1, and student 3. Fourth: student 2, student 3, and student 1. Fifth: student 3, student 2, and student 1. and finally, Sixth: student 3, student 1, and student 2. The question was, "How many ways can 3 students be selected from a group of 10 students?" Not How many ways can you CHOSE 3 from a group of 10 students?

  • @robertkelley3437
    @robertkelley3437 9 หลายเดือนก่อน

    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?

  • @pas6295
    @pas6295 9 หลายเดือนก่อน

    d is the answer

  • @CarmenRosas-ds5rn
    @CarmenRosas-ds5rn 8 หลายเดือนก่อน

    60

  • @bone0944
    @bone0944 9 หลายเดือนก่อน

    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.

  • @pas6295
    @pas6295 9 หลายเดือนก่อน

    120 ways

  • @hiteshkarthik4387
    @hiteshkarthik4387 9 หลายเดือนก่อน

    120

  • @dwaipayandattaroy9801
    @dwaipayandattaroy9801 9 หลายเดือนก่อน

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

  • @herbertklumpp2969
    @herbertklumpp2969 9 หลายเดือนก่อน

    10 over. 3

  • @nancyholmquist2690
    @nancyholmquist2690 9 หลายเดือนก่อน

    It does not make sense!

  • @twolery1514
    @twolery1514 9 หลายเดือนก่อน

    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.

  • @pamspencer5377
    @pamspencer5377 9 หลายเดือนก่อน

    A

  • @thomasschodt7691
    @thomasschodt7691 9 หลายเดือนก่อน

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

  • @user-qe8gw2xw1p
    @user-qe8gw2xw1p 9 หลายเดือนก่อน

    You’re explanation was too long

  • @homayounshirazi9550
    @homayounshirazi9550 9 หลายเดือนก่อน

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

  • @stevenk6638
    @stevenk6638 9 หลายเดือนก่อน

    your videos move WAY too slow !

  • @guywhyre2543
    @guywhyre2543 9 หลายเดือนก่อน

    @bulldog6925 below is CORRECT. My degree in Applied Mathematics tells me the answer is 10 X 9 X 8 = 720.

  • @bhumibhumi1470
    @bhumibhumi1470 8 หลายเดือนก่อน

    120

  • @CarmenRosas-ds5rn
    @CarmenRosas-ds5rn 8 หลายเดือนก่อน

    30