Harder Practice with Permutations and Combinations

Поделиться
HTML-код
  • Опубликовано: 4 апр 2018
  • Please comment if you have any questions or suggestions!

Комментарии • 40

  • @mariuszpopieluch7373
    @mariuszpopieluch7373 2 года назад +11

    Interesting problems but poor formulations and explanations.

  • @Brayden_Hoffart
    @Brayden_Hoffart Год назад

    Thanks! Helped a lot

  • @RG-tl5jl
    @RG-tl5jl 3 года назад +9

    We divided 4 for men because thats the society wants .
    Femenism Bruhhhh

  • @TALKmd
    @TALKmd 4 года назад +1

    Great

  • @emmanuelokpeh3259
    @emmanuelokpeh3259 2 года назад +4

    Great piece!!!. It would had been better if your questions are made legible by increasing the size of your character.

  • @simon-pierredandrea2928
    @simon-pierredandrea2928 Год назад

    Thanks

  • @noeltuplano6085
    @noeltuplano6085 3 года назад

    Great one!

  • @sk99gaming99
    @sk99gaming99 Год назад +1

    I solved the 2 question using quadraticn formula I got 1+-root 79/2

  • @Dhruvbala
    @Dhruvbala 2 года назад +6

    I'm still a little confused as to why you don't divide by 8 on the first one. Wouldn't it be a different arrangement if each individual shifted over by one?

    • @doublematter6132
      @doublematter6132 2 года назад +2

      Yeah that's what I wanted to ask

    • @sudiptabhaskor6060
      @sudiptabhaskor6060 2 года назад +12

      Think it this way. What if the table wasn't a circular one but a straight one? In that case, the number of possible arrangement would be 2*4!*4!*. The 2 accounts for the fact that you could start the arrangement either with a man or a woman. So 2 ways. Now when you think of a circular table, you have to consider Rotational Symmetry( That's the key term). For 8 people we have 8 rotational symmetry,right? So you actually divide up everything by 8. You have 144.

  • @narsinghyadav1071
    @narsinghyadav1071 3 месяца назад

    I like to think of the 1st question differently. From my what I learned about circular permutation, we can think of it as linear arrangement but we've overcounted by the possible rotations. We can rotate the table in such possible arrangement, and it happens 'n times for n seats'. I'd like check my understanding here, I'll be glad if anyone could help...
    So what I did was (for the 1st question)--
    First I considered linear arrangements, I came up with 2 possible cases - MenWomenMean...+ WomenMenWomen...
    And divided by 8 (8 seats)
    And got the same answer
    I just wanted to check if theres something wrong with my logic

    • @TragicGFuel
      @TragicGFuel 3 месяца назад

      How do you account for cases where the two supposed cases just happen to be the same?
      MWMW could be interpreted as WMWM based on the starting point

  • @Markjohn26
    @Markjohn26 3 года назад

    It help me thanks

  • @SoFia-fd6qj
    @SoFia-fd6qj 3 года назад

    May i ask what category does number 1 belongs to?? Is it a permutation or a combination?

    • @ogkakulalo3872
      @ogkakulalo3872 3 года назад +1

      permutation

    • @Iseeyou0165
      @Iseeyou0165 3 года назад

      permutation as the order matters (men and female needs to be alternate to each other)!

  • @kathleenolivier1588
    @kathleenolivier1588 3 года назад +3

    Hi sir or anyone can you tell me the answer for this question.
    A coach has 16 players and can pick from 11 players for a match. it consist of 7 specialist batsmen, 4 fast bowlers, 3 spinner and 2 wicketkeeper. how many different teams can be formed if it must comprised of 6 specialist batsmen, 3 fast bowlers , 1 spinner and a wicketkeeper.

    • @oliwood1592
      @oliwood1592 3 года назад +8

      7C6 (batsmen) x 4C3 (bowlers) x 3C1 (spinners) x 2C1 (wicketkeepers) = 168

    • @HHH21
      @HHH21 2 года назад

      More accurate question:
      A coach need 11 players and can pick from 16 players for a match.

    • @billlu1076
      @billlu1076 2 года назад

      @@HHH21 16C11

    • @mynameisjeff9124
      @mynameisjeff9124 2 года назад

      @@HHH21 No, that's wrong. The correct one is the answer by @Oli Wood, because if you choose 11 players from 16, then this also has the possibility to pick all 7 specialist batsmen and all 4 fast bowlers and one of the others.

    • @HHH21
      @HHH21 2 года назад

      @@mynameisjeff9124 How can you pick from only 11 players if total are 16?
      And there isn't any possibility to pick all 7 batsman and more because restrictions are already given

  • @danielfrt
    @danielfrt 4 месяца назад

    Second problem is not right, we can prove it easily by stating the problem as: A bookshelf has 5 fiction books and 6 non-fictioon books. In how many ways can we choose two books of each type? the answer is not 150. the equation for the original question is simply 6x=150

    • @ZoeZ.U.
      @ZoeZ.U. 20 дней назад

      The answer in the vedio is correct. There are 5 non fiction books. That equation you're talking about is wrong, because what is x supposed to represent

  • @bingw4416
    @bingw4416 7 месяцев назад

    wu

  • @srivatsava0605
    @srivatsava0605 4 года назад +1

    I didnt get the 3rd one tho!

    • @gnumani
      @gnumani 4 года назад +1

      Ik o bajge

    • @danielknutson8238
      @danielknutson8238 3 года назад +1

      Im pretty sure that the 3rd one he calculated the odds of getting a hand all the same colour and not the # of possibilities

    • @prakharpant1711
      @prakharpant1711 3 года назад

      @@danielknutson8238 yeah he did that too but if you saw the video he first found out the number of possibilities, then took out the probability. Number of possibilities are 2x26C5 and probability is just that divided by all possibilities i.e. 52C5. These we get by the basic combination formula of choosing r things out of n distinct things. What I don’t understand is the first one. There’s something called the 8 way symmetry of a circle and if you think about it. 4! X 4! should be divided by 8 because all 8 places are identical in a circle but it’s not. Why is that?

    • @ZoeZ.U.
      @ZoeZ.U. 20 дней назад

      ​@@danielknutson8238 wasnt the question asking frlor how many HANDS have cards all the same colour? So why did he solve it like that. Wouldnt u need to divide the 26 by 5

    • @dog360
      @dog360 День назад

      ​@@ZoeZ.U. he definitely misinterpreted the question

  • @aspektstv3149
    @aspektstv3149 2 года назад +4

    You call this hard practice??? What's easy then??

    • @ZoeZ.U.
      @ZoeZ.U. 20 дней назад

      I didnt understand the first one

    • @ZoeZ.U.
      @ZoeZ.U. 20 дней назад

      Or third

  • @shibajyotichoudhury8186
    @shibajyotichoudhury8186 4 года назад +6

    These are not hard at all....easy peasy