From a deck of 52 cards, extract 10. In how many combinations do you get at least one ace?












0














If from a deck of 52 cards, I extract 10. In how many combinations do you get at least one ace?



I have come up with two possible answers, but I don't know which one is the right one and why.



So one is $$4{51 choose 9}$$
the reasoning is that first I extract an ace from the four that there are, and then I have ${51 choose 9}$ combinations for the other 9 cards.



The second is
$${52 choose 10} - {48 choose 10}$$ reasoning that there are ${52 choose 10}$ total combinations of ten cards and I subtract ${48 choose 10}$ combinations without any aces.



So from other questions it seems that the first one is the correct answer, but why would the second one be wrong?



Thanks for the answers.






combinatorics combinations







share|cite|improve this question




















  • 3




    Your first method counts hands with more than one ace multiple times.
    – lulu
    Nov 27 '18 at 16:49










  • @lulu That's important as a point, but the question wants to count all hands with at least one ace. There is still overcounting going on though.
    – amWhy
    Nov 27 '18 at 16:50












  • @amWhy lulu is pointing out that certain hands are counted too many times.
    – Théophile
    Nov 27 '18 at 16:53






  • 2




    The second method is correct. It is the best way for the general question. - to obtain probability of at least one, calculate the probability of none and subtract from $1$.
    – herb steinberg
    Nov 27 '18 at 17:00






  • 2




    @amWhy Not sure I see your point. if ${X_i}_{i=1}^8$ denote specific non-aces, then the OP's method counts $Aspadesuit, Aheartsuit, {X_i}_{i=1}^8$ as one case, and the equivalent reordered hand $Aheartsuit, Aspadesuit, {X_i}_{i=1}^8$ as another. I understand the OP is counting hands containing at least one ace.
    – lulu
    Nov 27 '18 at 17:01


















0














If from a deck of 52 cards, I extract 10. In how many combinations do you get at least one ace?



I have come up with two possible answers, but I don't know which one is the right one and why.



So one is $$4{51 choose 9}$$
the reasoning is that first I extract an ace from the four that there are, and then I have ${51 choose 9}$ combinations for the other 9 cards.



The second is
$${52 choose 10} - {48 choose 10}$$ reasoning that there are ${52 choose 10}$ total combinations of ten cards and I subtract ${48 choose 10}$ combinations without any aces.



So from other questions it seems that the first one is the correct answer, but why would the second one be wrong?



Thanks for the answers.






combinatorics combinations







share|cite|improve this question




















  • 3




    Your first method counts hands with more than one ace multiple times.
    – lulu
    Nov 27 '18 at 16:49










  • @lulu That's important as a point, but the question wants to count all hands with at least one ace. There is still overcounting going on though.
    – amWhy
    Nov 27 '18 at 16:50












  • @amWhy lulu is pointing out that certain hands are counted too many times.
    – Théophile
    Nov 27 '18 at 16:53






  • 2




    The second method is correct. It is the best way for the general question. - to obtain probability of at least one, calculate the probability of none and subtract from $1$.
    – herb steinberg
    Nov 27 '18 at 17:00






  • 2




    @amWhy Not sure I see your point. if ${X_i}_{i=1}^8$ denote specific non-aces, then the OP's method counts $Aspadesuit, Aheartsuit, {X_i}_{i=1}^8$ as one case, and the equivalent reordered hand $Aheartsuit, Aspadesuit, {X_i}_{i=1}^8$ as another. I understand the OP is counting hands containing at least one ace.
    – lulu
    Nov 27 '18 at 17:01
















0












0








0







If from a deck of 52 cards, I extract 10. In how many combinations do you get at least one ace?



I have come up with two possible answers, but I don't know which one is the right one and why.



So one is $$4{51 choose 9}$$
the reasoning is that first I extract an ace from the four that there are, and then I have ${51 choose 9}$ combinations for the other 9 cards.



The second is
$${52 choose 10} - {48 choose 10}$$ reasoning that there are ${52 choose 10}$ total combinations of ten cards and I subtract ${48 choose 10}$ combinations without any aces.



So from other questions it seems that the first one is the correct answer, but why would the second one be wrong?



Thanks for the answers.






combinatorics combinations







share|cite|improve this question















If from a deck of 52 cards, I extract 10. In how many combinations do you get at least one ace?



I have come up with two possible answers, but I don't know which one is the right one and why.



So one is $$4{51 choose 9}$$
the reasoning is that first I extract an ace from the four that there are, and then I have ${51 choose 9}$ combinations for the other 9 cards.



The second is
$${52 choose 10} - {48 choose 10}$$ reasoning that there are ${52 choose 10}$ total combinations of ten cards and I subtract ${48 choose 10}$ combinations without any aces.



So from other questions it seems that the first one is the correct answer, but why would the second one be wrong?



Thanks for the answers.






combinatorics combinations




combinatorics combinations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 '18 at 23:56









N. F. Taussig

43.5k93355




43.5k93355










asked Nov 27 '18 at 16:45









Miguel Vázquez Caraballo

41




41








  • 3




    Your first method counts hands with more than one ace multiple times.
    – lulu
    Nov 27 '18 at 16:49










  • @lulu That's important as a point, but the question wants to count all hands with at least one ace. There is still overcounting going on though.
    – amWhy
    Nov 27 '18 at 16:50












  • @amWhy lulu is pointing out that certain hands are counted too many times.
    – Théophile
    Nov 27 '18 at 16:53






  • 2




    The second method is correct. It is the best way for the general question. - to obtain probability of at least one, calculate the probability of none and subtract from $1$.
    – herb steinberg
    Nov 27 '18 at 17:00






  • 2




    @amWhy Not sure I see your point. if ${X_i}_{i=1}^8$ denote specific non-aces, then the OP's method counts $Aspadesuit, Aheartsuit, {X_i}_{i=1}^8$ as one case, and the equivalent reordered hand $Aheartsuit, Aspadesuit, {X_i}_{i=1}^8$ as another. I understand the OP is counting hands containing at least one ace.
    – lulu
    Nov 27 '18 at 17:01
















  • 3




    Your first method counts hands with more than one ace multiple times.
    – lulu
    Nov 27 '18 at 16:49










  • @lulu That's important as a point, but the question wants to count all hands with at least one ace. There is still overcounting going on though.
    – amWhy
    Nov 27 '18 at 16:50












  • @amWhy lulu is pointing out that certain hands are counted too many times.
    – Théophile
    Nov 27 '18 at 16:53






  • 2




    The second method is correct. It is the best way for the general question. - to obtain probability of at least one, calculate the probability of none and subtract from $1$.
    – herb steinberg
    Nov 27 '18 at 17:00






  • 2




    @amWhy Not sure I see your point. if ${X_i}_{i=1}^8$ denote specific non-aces, then the OP's method counts $Aspadesuit, Aheartsuit, {X_i}_{i=1}^8$ as one case, and the equivalent reordered hand $Aheartsuit, Aspadesuit, {X_i}_{i=1}^8$ as another. I understand the OP is counting hands containing at least one ace.
    – lulu
    Nov 27 '18 at 17:01










3




3




Your first method counts hands with more than one ace multiple times.
– lulu
Nov 27 '18 at 16:49




Your first method counts hands with more than one ace multiple times.
– lulu
Nov 27 '18 at 16:49












@lulu That's important as a point, but the question wants to count all hands with at least one ace. There is still overcounting going on though.
– amWhy
Nov 27 '18 at 16:50






@lulu That's important as a point, but the question wants to count all hands with at least one ace. There is still overcounting going on though.
– amWhy
Nov 27 '18 at 16:50














@amWhy lulu is pointing out that certain hands are counted too many times.
– Théophile
Nov 27 '18 at 16:53




@amWhy lulu is pointing out that certain hands are counted too many times.
– Théophile
Nov 27 '18 at 16:53




2




2




The second method is correct. It is the best way for the general question. - to obtain probability of at least one, calculate the probability of none and subtract from $1$.
– herb steinberg
Nov 27 '18 at 17:00




The second method is correct. It is the best way for the general question. - to obtain probability of at least one, calculate the probability of none and subtract from $1$.
– herb steinberg
Nov 27 '18 at 17:00




2




2




@amWhy Not sure I see your point. if ${X_i}_{i=1}^8$ denote specific non-aces, then the OP's method counts $Aspadesuit, Aheartsuit, {X_i}_{i=1}^8$ as one case, and the equivalent reordered hand $Aheartsuit, Aspadesuit, {X_i}_{i=1}^8$ as another. I understand the OP is counting hands containing at least one ace.
– lulu
Nov 27 '18 at 17:01






@amWhy Not sure I see your point. if ${X_i}_{i=1}^8$ denote specific non-aces, then the OP's method counts $Aspadesuit, Aheartsuit, {X_i}_{i=1}^8$ as one case, and the equivalent reordered hand $Aheartsuit, Aspadesuit, {X_i}_{i=1}^8$ as another. I understand the OP is counting hands containing at least one ace.
– lulu
Nov 27 '18 at 17:01












2 Answers
2






active

oldest

votes


















2














The first approach needs to be modified to ${4choose1}{48choose 9}+{4choose2}{48choose 8}+{4choose 3}{48choose 7}+{4choose 4}{48choose 6}$, for instance.



The second is correct.






share|cite|improve this answer





























    0














    To see why your first answer is not correct, assume you have a deck of 3 with 2 aces and you'll draw 2 cards with at least one ace. Call the cards $A_1,A_2,B$



    Based on your logic, you compute $2 {2 choose 1}=4$ combinations. However, clearly you have only $3$. This is due to double counting ${A_1,A_2}$






    share|cite|improve this answer





















      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3015999%2ffrom-a-deck-of-52-cards-extract-10-in-how-many-combinations-do-you-get-at-leas%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2














      The first approach needs to be modified to ${4choose1}{48choose 9}+{4choose2}{48choose 8}+{4choose 3}{48choose 7}+{4choose 4}{48choose 6}$, for instance.



      The second is correct.






      share|cite|improve this answer


























        2














        The first approach needs to be modified to ${4choose1}{48choose 9}+{4choose2}{48choose 8}+{4choose 3}{48choose 7}+{4choose 4}{48choose 6}$, for instance.



        The second is correct.






        share|cite|improve this answer
























          2












          2








          2






          The first approach needs to be modified to ${4choose1}{48choose 9}+{4choose2}{48choose 8}+{4choose 3}{48choose 7}+{4choose 4}{48choose 6}$, for instance.



          The second is correct.






          share|cite|improve this answer












          The first approach needs to be modified to ${4choose1}{48choose 9}+{4choose2}{48choose 8}+{4choose 3}{48choose 7}+{4choose 4}{48choose 6}$, for instance.



          The second is correct.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 27 '18 at 23:19









          Chris Custer

          10.8k3724




          10.8k3724























              0














              To see why your first answer is not correct, assume you have a deck of 3 with 2 aces and you'll draw 2 cards with at least one ace. Call the cards $A_1,A_2,B$



              Based on your logic, you compute $2 {2 choose 1}=4$ combinations. However, clearly you have only $3$. This is due to double counting ${A_1,A_2}$






              share|cite|improve this answer


























                0














                To see why your first answer is not correct, assume you have a deck of 3 with 2 aces and you'll draw 2 cards with at least one ace. Call the cards $A_1,A_2,B$



                Based on your logic, you compute $2 {2 choose 1}=4$ combinations. However, clearly you have only $3$. This is due to double counting ${A_1,A_2}$






                share|cite|improve this answer
























                  0












                  0








                  0






                  To see why your first answer is not correct, assume you have a deck of 3 with 2 aces and you'll draw 2 cards with at least one ace. Call the cards $A_1,A_2,B$



                  Based on your logic, you compute $2 {2 choose 1}=4$ combinations. However, clearly you have only $3$. This is due to double counting ${A_1,A_2}$






                  share|cite|improve this answer












                  To see why your first answer is not correct, assume you have a deck of 3 with 2 aces and you'll draw 2 cards with at least one ace. Call the cards $A_1,A_2,B$



                  Based on your logic, you compute $2 {2 choose 1}=4$ combinations. However, clearly you have only $3$. This is due to double counting ${A_1,A_2}$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 27 '18 at 17:10









                  karakfa

                  1,973811




                  1,973811






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.





                      Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                      Please pay close attention to the following guidance:


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3015999%2ffrom-a-deck-of-52-cards-extract-10-in-how-many-combinations-do-you-get-at-leas%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      Bundesstraße 106

                      Image
                      Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück ...
                      Read more

                      Verónica Boquete

                      Image
                      Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl ...
                      Read more

                      Ida-Boy-Ed-Garten

                      Image
                      Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich ...
                      Read more