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The number of ways that a mixed doubles tennis game can be arranged from 7 married couples if no husband and wife play on the same game.

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Probably explaining it is better......You have to pick 2 people of same gender and 2 of other gender so that you have 4 people for mixed doubles. To avoid same wife-husband combo, pick 4 pairs among 7 pairs and see that 2 wifes and 2 husbands play together (In such case, there won't be a same wife-husband...
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Probably explaining it is better......You have to pick 2 people of same gender and 2 of other gender so that you have 4 people for mixed doubles. To avoid same wife-husband combo, pick 4 pairs among 7 pairs and see that 2 wifes and 2 husbands play together (In such case, there won't be a same wife-husband combo). The way you could do it is 7p4. Half it, to avoid repetitions. you get the number 420. read less
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The first man can be selected in 7 ways. the first woman in 6 ways. the second man should not be the husband of the woman. He can be selected in 5 ways(excluding the first man and the the first woman's hus). the second woman can be selected in 4ways(excluding the first woman and the wives of the 2 men). Total...
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The first man can be selected in 7 ways. the first woman in 6 ways. the second man should not be the husband of the woman. He can be selected in 5 ways(excluding the first man and the the first woman's hus). the second woman can be selected in 4ways(excluding the first woman and the wives of the 2 men). Total selections = 7x6x5x4 = 840 arrangements = 840x4x2x2 = 13440. read less
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Total male members can be selected in 7C2 Ways = 21 , Two female members from the other couple can be selected in 5C2 ways = 10 , and two male and two female can be arranged among themselves in 2 ways. there fore total is = 21*10* 2 = 420.
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Te first man can be selected in 7 ways. the first woman in 6 ways. the second man should not be the husband of the woman. He can be selected in 5 ways(excluding the first man and the the first woman's hus). the second woman can be selected in 4ways(excluding the first woman and the wives of the 2 men). Total...
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Te first man can be selected in 7 ways. the first woman in 6 ways. the second man should not be the husband of the woman. He can be selected in 5 ways(excluding the first man and the the first woman's hus). the second woman can be selected in 4ways(excluding the first woman and the wives of the 2 men). Total selections = 7x6x5x4 = 840 arrangements = 840x4x2x2 = 13440. read less
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No. of ways to select 1st team 7 ways to select a male and then 6 ways to select a female= 7*6 = 42 No. of ways to select 2nd team 5 ways to select a male (1 male is selected reducing the options to 6 and since one female is selected which should not be the wife, the options are reduced to 5) and...
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No. of ways to select 1st team 7 ways to select a male and then 6 ways to select a female= 7*6 = 42 No. of ways to select 2nd team 5 ways to select a male (1 male is selected reducing the options to 6 and since one female is selected which should not be the wife, the options are reduced to 5) and then 4 ways to select female (1 female is selected reducing the options to 6 and since 2 males have been selected who should not be the husband, the options are reduced to 4) = 5*4 = 20 Total no. of ways = 42 * 20 = 840 Now the only thing left is to remove the duplication i.e. team A vs B and team B vs A is the same combination. To remove this duplication, divide the result by 2. Hence answer would be 420. read less
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i believe in authantic teaching. basic concepts of a child makes the subject easy and intresting. i am working on this feild for last ten years and got very good results children who dare with maths get intrest in it and perform excelent in their exams maths become fun for them

7 couples = 14 members No. of spaces to be filled in mixed doubles game = 4 The first space can be filled with any of the 14 members. The second space cannot be filled with the previous person and partner of that person(to avoid husband wife combo). No. of people who can be selected for 2nd space are...
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7 couples = 14 members No. of spaces to be filled in mixed doubles game = 4 The first space can be filled with any of the 14 members. The second space cannot be filled with the previous person and partner of that person(to avoid husband wife combo). No. of people who can be selected for 2nd space are 12. Similarly third space can be filled by 10 individuals and forth one by 8. No. of ways of selection are 14*12*10*8= 13440 read less
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Math Simplified

Hi, My answer for the question: for question The number of ways that a mixed doubles tennis game can be arranged from 7 married couples if no husband and wife play on the same game. Ans: In mixed double we have four position. 1. First position of a man can be filled with 7 ways as we can...
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Hi, My answer for the question: for question The number of ways that a mixed doubles tennis game can be arranged from 7 married couples if no husband and wife play on the same game. Ans: In mixed double we have four position. 1. First position of a man can be filled with 7 ways as we can select any man out of seven. --- 7 2. Second position we need to select woman and we can select any 6 woman out of seven as we can not select husband of the person at First position. -- 6 3. In third position we can select any one person out of 5. Why because we should have any couple playing the game any of the side. After first position we are remaining with six man, out of which one would be the husband of second position woman, so we have 5 ways for third position. -- 5 4. For fourth position we can select any one woman out of 4. As after second position is filled we are left with five woman. We can select the woman who is the wife of third position Man. So we are left with 4 woman at fourth position. --4 So total possible arrangement would be all combination of these four position : 7*6*5*4=840. Thanks, Amit read less
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