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trouble wid P&C quest

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trouble wid P&C quest

by willbeatthegmat » Fri Nov 07, 2008 2:12 am
Q-1) There are 5 married couples and a group of three is to be formed out of them; how many arrangements are there if a husband and wife may not be in the same group?
Ans. 80

Q-2) How many 3-digit numbers are possible using the digits 0, 1, 4, 5, 7 and 9?
Ans: 180, shud b 100

Q-3) Find the number of combinations of 7 objects taking 2 at a time.
Ans-21

Q-4) How many ways can 7 people be arranged around a round table?
Ans 720
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Re: trouble wid P&C quest

by logitech » Fri Nov 07, 2008 3:12 am
willbeatthegmat wrote:
Q-1) There are 5 married couples and a group of three is to be formed out of them; how many arrangements are there if a husband and wife may not be in the same group?
Ans. 80

10x8x6 different ways

but since group of three can be arranged in 3x2x1 = 6 ways

10x8x6 / 6 = 80

Q-2) How many 3-digit numbers are possible using the digits 0, 1, 4, 5, 7 and 9?
Ans: 180, shud b 100

5x5x4 = 100

180 seems wrong

Q-3) Find the number of combinations of 7 objects taking 2 at a time.
Ans-21

7!/(5!2!) = 21

Q-4) How many ways can 7 people be arranged around a round table?
Ans 720
Round table combinations:

(n-1)!

(7-1)! = 6! = 720
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by willbeatthegmat » Fri Nov 07, 2008 3:37 am
y did u subtract 1 from it...pls elaborate
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by willbeatthegmat » Fri Nov 07, 2008 3:38 am
y did u subtract 1 from it...pls elaborate
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by jimmiejaz » Fri Nov 07, 2008 3:39 am
Hi logitech,

In the 2nd question. it doesnt say that repetition of the digits are not allowed.
so i guess that is what makes the difference...
If repetition is allowed, we get 5*6*6 = 180
If repetition is not allowed, we get 5*5*4 = 100

Hope it helps...
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by jimmiejaz » Fri Nov 07, 2008 3:41 am
willbeatthegmat wrote:y did u subtract 1 from it...pls elaborate
for round table or we call it circular permutations, the no of arrangements is (n-1)!.
eg if u want to arrange 2 guys in a round table, how many arrangements you have?
U can arrange them in one way. coz anticlockwise or clockwise makes no difference. If it does, then it is same as a row.
Hope it helps.
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by willbeatthegmat » Fri Nov 07, 2008 4:49 am
jimmiejaz wrote:Hi logitech,

In the 2nd question. it doesnt say that repetition of the digits are not allowed.
so i guess that is what makes the difference...
If repetition is allowed, we get 5*6*6 = 180
If repetition is not allowed, we get 5*5*4 = 100

Hope it helps...
appreciate ur response jimmi...ma approach in solving dis quest was i think a bit different n cudn't understand d way u solved..
i took 6*5*4 =120 coz we have 3 positions. And since we cannot consider number startin wid 0 as three digits...i subtracted 1/6 of 120 frm 120 n got 100 as d answer. Am i missin sumthin?..n is this d right approach?
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by jimmiejaz » Fri Nov 07, 2008 4:55 am
willbeatthegmat wrote:
jimmiejaz wrote:Hi logitech,

In the 2nd question. it doesnt say that repetition of the digits are not allowed.
so i guess that is what makes the difference...
If repetition is allowed, we get 5*6*6 = 180
If repetition is not allowed, we get 5*5*4 = 100

Hope it helps...
appreciate ur response jimmi...ma approach in solving dis quest was i think a bit different n cudn't understand d way u solved..
i took 6*5*4 =120 coz we have 3 positions. And since we cannot consider number startin wid 0 as three digits...i subtracted 1/6 of 120 frm 120 n got 100 as d answer. Am i missin sumthin?..n is this d right approach?
according to ur approach....
again u supposed repetition is not allowed.....
so from ur approach, no of ways = 6*6*6=216
subtract 1/6 ir 36. 216-36=180....
Hope it clarifies ur confusion....
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by logitech » Fri Nov 07, 2008 10:40 am
jimmiejaz rocks!!
LGTCH
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by jimmiejaz » Fri Nov 07, 2008 10:49 am
Hi Logitech,

thanks!!!! If you can please solve this question... will be a great help!!!!!

https://www.beatthegmat.com/probability- ... 21551.html
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by cramya » Fri Nov 07, 2008 4:53 pm
There are 5 married couples and a group of three is to be formed out of them; how many arrangements are there if a husband and wife may not be in the same group?
Lets do this without any constraints:

Selecting 3 people out of 10

10C3 = 120

Lets find out the number of groups in which the married couples will exist

H1 W1 H2 W2 H3 W3 H4 W4 H5 W5

Lets take just H1 and W1 groups

H1 W1
--- --- ----

The third spot can be taken by any of the remaining 8 people so 8 such groups exist. For 5 married couple 5 * 8 =40 such group exists

So we subtract this 40(number of groups in which the married couples will exist) from 120(Selecting 3 people out of 10) to get us to the groups with no married couples in them i.e 80


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by logitech » Fri Nov 07, 2008 5:22 pm
cramya wrote:



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It is okay to say THANK YOU Cramya.. :lol:
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