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A group of 10 people consists of 3 married couples and 4 [#permalink]

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17 May 2011, 08:18

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A group of 10 people consists of 3 married couples and 4 single men. A committee of 4 is to be from the 10 people. How many different committees can be formed if the committee can consist of at most 1 married couple?

Can someone employ the forward method i.e. (not the 1-x method to deduce the answer) ? I think I over counting as my answer overshoots the OA

Answer = No of ways to select one couple and two singles + No of ways to select all singles

Using 1 - x method the question is piece of cake. But if you employ the forward method i.e calculating the combinations which have a couple or all singles, the complexity jumps up.

1 - x method The number of ways = 10C4 - (no of ways to select 2 couple) = 10C4 - 3C2 = 207

Forward method --------------- No of ways to select one couple and two singles + No of ways to select all singles

Obvious cases - All 4 singles & 1 married couple and 2 singles

Other cases - 1 marrried person from a couple and 3 singles 1 married couple, 1 married person from a couple and 1 single 1 married couple, 2 married person from different couples 1 married person from each couple and 1 single 1 married person from 2 separate couples and 2 singles

A group of 10 people consists of 3 married couples and 4 single men. A committee of 4 is to be from the 10 people. How many different committees can be formed if the committee can consist of at most 1 married couple?

Can someone employ the forward method i.e. (not the 1-x method to deduce the answer) ? I think I over counting as my answer overshoots the OA

Answer = No of ways to select one couple and two singles + No of ways to select all singles

Using 1 - x method the question is piece of cake. But if you employ the forward method i.e calculating the combinations which have a couple or all singles, the complexity jumps up.

1 - x method The number of ways = 10C4 - (no of ways to select 2 couple) = 10C4 - 3C2 = 207

Forward method --------------- No of ways to select one couple and two singles + No of ways to select all singles

Let A----B C----D E ----F are couples and 1 2 3 4 are singles

1st scenario .. 4 singles out of 4 singles ------- 4c4 = 1 2nd scenario 2 Married singles + 2 singles ------ ( 6*4)/2 * 4c2 = 72 3rd 1 Married single + 3 singles ------- 6*4c3 = 24 4th 3 married single + 1 single ------- (6*4*2)/3!*4C1 = 32 5th scenario 1Couple + 1 Married single + 1 single ---- 3c1*4*4c1= 48 6th scenario 1 couple + 2 Married singles ------- 3c1*( 4*2)/2 = 12 7th scenario 1 couple + 2 singles --------- 3c1*4C2 = 18

A group of 10 people consists of 3 married couples and 4 single men. A committee of 4 is to be from the 10 people. How many different committees can be formed if the committee can consist of at most 1 married couple?

Can someone employ the forward method i.e. (not the 1-x method to deduce the answer) ? I think I over counting as my answer overshoots the OA

Answer = No of ways to select one couple and two singles + No of ways to select all singles

Using 1 - x method the question is piece of cake. But if you employ the forward method i.e calculating the combinations which have a couple or all singles, the complexity jumps up.

1 - x method The number of ways = 10C4 - (no of ways to select 2 couple) = 10C4 - 3C2 = 207

Forward method --------------- No of ways to select one couple and two singles + No of ways to select all singles

Let A----B C----D E ----F are couples and 1 2 3 4 are singles

1st scenario .. 4 singles out of 4 singles ------- 4c4 = 1 2nd scenario 2 Married singles + 2 singles ------ ( 6*4)/2 * 4c2 = 72 3rd 1 Married single + 3 singles ------- 6*4c3 = 24 4th 3 married single + 1 single ------- (6*4*2)/3!*4C1 = 32 5th scenario 1Couple + 1 Married single + 1 single ---- 3c1*4*4c1= 48 6th scenario 1 couple + 2 Married singles ------- 3c1*( 4*2)/2 = 12 7th scenario 1 couple + 2 singles --------- 3c1*4C2 = 18

Total = 1+72+24+ 32+48+12+18 = 207

hi - you could have simply done this ..

Total number of ways of selecting 4 people out of 10 = 10C4 Total number of ways of selecting 2 married couple = 3C2

total number of combinations 10c4 = 10!/4!6! = 210 total number of married couple = 3c2 = 3!/2! = 3 total number of possible commities are 207 _________________

We can use the other way of thinking, that is (total possibility of selecting 4 people)-(selecting two couples) then the answer will be 10C4-3C2=210-3=207. That's much more quickly

One way to do problem: Answer = No of ways to select one couple and two singles + No of ways to select all singles=207=official answer

How about the 1-x method

"The number of ways = 10C4 - (no of ways to select 2 couple) = 10C4 - 3C2 = 207"

Since when has the opposite of "at most 1" equaled to exact no of ways to select 2 couples? Could someone help me grasp why this is conceptually the opposite of exactly 1 couple? _________________

One way to do problem: Answer = No of ways to select one couple and two singles + No of ways to select all singles=207=official answer

How about the 1-x method

"The number of ways = 10C4 - (no of ways to select 2 couple) = 10C4 - 3C2 = 207"

Since when has the opposite of "at most 1" equaled to exact no of ways to select 2 couples? Could someone help me grasp why this is conceptually the opposite of exactly 1 couple?

You need to select a committee of 4. You can do it in 3 ways: a. Select no couple and 4 others b. Select one couple and 2 others c. Select two couples

'At most one' means either no couples or only one couple. What is the complement of at most one? The only other way in which you can select a committee of 4 is by selecting 2 couples. So you can find the number of ways of selecting 2 couples and subtract that from the total number of ways of selecting 4 people. You will get the number of ways in which there will be at most one couple. _________________

One way to do problem: Answer = No of ways to select one couple and two singles + No of ways to select all singles=207=official answer

How about the 1-x method

"The number of ways = 10C4 - (no of ways to select 2 couple) = 10C4 - 3C2 = 207"

Since when has the opposite of "at most 1" equaled to exact no of ways to select 2 couples? Could someone help me grasp why this is conceptually the opposite of exactly 1 couple?

You need to select a committee of 4. You can do it in 3 ways: a. Select no couple and 4 others b. Select one couple and 2 others c. Select two couples

'At most one' means either no couples or only one couple. What is the complement of at most one? The only other way in which you can select a committee of 4 is by selecting 2 couples. So you can find the number of ways of selecting 2 couples and subtract that from the total number of ways of selecting 4 people. You will get the number of ways in which there will be at most one couple.

Namaste aur shukria so much again Karishma. Made it clear within the context of the problem _________________

Re: A group of 10 people consists of 3 married couples and 4 [#permalink]

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16 Feb 2013, 04:10

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gmat1220 wrote:

A group of 10 people consists of 3 married couples and 4 single men. A committee of 4 is to be from the 10 people. How many different committees can be formed if the committee can consist of at most 1 married couple?

Can someone employ the forward method i.e. (not the 1-x method to deduce the answer) ? I think I over counting as my answer overshoots the OA

Answer = No of ways to select one couple and two singles + No of ways to select all singles

Using 1 - x method the question is piece of cake. But if you employ the forward method i.e calculating the combinations which have a couple or all singles, the complexity jumps up.

1 - x method The number of ways = 10C4 - (no of ways to select 2 couple) = 10C4 - 3C2 = 207

Forward method --------------- No of ways to select one couple and two singles + No of ways to select all singles

Re: A group of 10 people consists of 3 married couples and 4 [#permalink]

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22 Feb 2013, 07:47

my question is : i understand we can do like selecting 2 couples and subtract that from the total number of ways of selecting 4 people. now there 6 married peple and 4 singles so it should be 10c4 - 6x5x4x3/4x3x2x1

but i dont understand how 10c4- 3x2/2 comes?? can someone please explain. _________________

"Giving kudos" is a decent way to say "Thanks" and motivate contributors. Please use them, it won't cost you anything

Re: A group of 10 people consists of 3 married couples and 4 [#permalink]

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22 Feb 2013, 08:04

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skamal7 wrote:

my question is : i understand we can do like selecting 2 couples and subtract that from the total number of ways of selecting 4 people. now there 6 married peple and 4 singles so it should be 10c4 - 6x5x4x3/4x3x2x1

but i dont understand how 10c4- 3x2/2 comes?? can someone please explain.

We have 3 couples, so 3C2=3!/2!=3 is the number of ways to choose 2 couples out of 3.

Re: A group of 10 people consists of 3 married couples and 4 [#permalink]

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22 Feb 2013, 08:10

bunnel thanks for kind response.

One thing i dont understand is while taking total number of ways of selecting 4 people we count the 3 married couples as 6 right.?so when we are selecting 2 married couples it should be 6x5x4x3/4! na??but why when selecting 2 married couples we are taking count just as 3x2/2!?? please do explain!!! _________________

"Giving kudos" is a decent way to say "Thanks" and motivate contributors. Please use them, it won't cost you anything

Re: A group of 10 people consists of 3 married couples and 4 [#permalink]

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22 Feb 2013, 08:13

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skamal7 wrote:

bunnel thanks for kind response.

One thing i dont understand is while taking total number of ways of selecting 4 people we count the 3 married couples as 6 right.?so when we are selecting 2 married couples it should be 6x5x4x3/4! na??but why when selecting 2 married couples we are taking count just as 3x2/2!?? please do explain!!!

We are interested in number of ways we can select 2 couples out of 3, so 2 entities out of 3. _________________

Re: A group of 10 people consists of 3 married couples and 4 [#permalink]

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13 Aug 2014, 08:23

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