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1 1+X+X^2+X^3+X^4+X^5<1/(1-X)? a. X>0 b. X<1 2 4

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CEO
Joined: 15 Aug 2003
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1 1+X+X^2+X^3+X^4+X^5<1/(1-X)? a. X>0 b. X<1 2 4 [#permalink]

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22 Sep 2003, 14:20
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1 1+X+X^2+X^3+X^4+X^5<1/(1-X)?
a. X>0
b. X<1

2 4 seniors,6 juniors,how many groups can be formed to include 3
persons within which there is at least 1 senior?(2 groups are
considered different so long as 1 group is different)

Kudos [?]: 919 [0], given: 781

Manager
Joined: 15 Sep 2003
Posts: 73

Kudos [?]: 10 [0], given: 0

Location: california

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22 Sep 2003, 15:25
ok...i'll give it a shot...

1) I get A (although 1/0 is undefined) i don't know if it's a typo or what, but x>=1 is sufficient, and x<1 isn't

2) i get C(10,3) - C(6,3)

total ways - all juniors

am i close??

Kudos [?]: 10 [0], given: 0

Intern
Joined: 03 May 2003
Posts: 28

Kudos [?]: 1 [0], given: 0

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22 Sep 2003, 17:13
2 4 seniors,6 juniors,how many groups can be formed to include 3
persons within which there is at least 1 senior?(2 groups are
considered different so long as 1 group is different)

Ok form a group of 3 with at least 1 senior
You can have SSS, SSJ, SJJ

SSS
4C3 = 4

SSJ
4C2 * 6C1 = 35

Sjj
4C1 * 6C2 = 60

Total ways = 4 + 35 +60 = 99

?

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Senior Manager
Joined: 22 Aug 2003
Posts: 257

Kudos [?]: 13 [0], given: 0

Location: Bangalore

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23 Sep 2003, 02:41
1+X+X^2+X^3+X^4+X^5 = (1-X^6)/(1-X)
Now, (1-X^6)/(1-X) < 1/(1-X)

1) X>0 Not suffecient.
2) X<1 Suffecient

2) Number of ways to form groups with 3 poeple such that atleast 1 is senior = 4C1*9C2 = 144

Kudos [?]: 13 [0], given: 0

SVP
Joined: 03 Feb 2003
Posts: 1603

Kudos [?]: 304 [0], given: 0

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23 Sep 2003, 03:25
is B sufficient for your inequality?

X=0 => 1<1 wrong
X=-1 => 0<1/2 right

your second solution scares me... what does 9C2 mean? where does it come from?

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Senior Manager
Joined: 22 Aug 2003
Posts: 257

Kudos [?]: 13 [0], given: 0

Location: Bangalore

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23 Sep 2003, 04:06
oh man.. silly mistake...
thanks for correction.
Second solution:
4C1 = number of ways one senior can be selcted
9C2 = number of ways any 2 people can be selected from remaining 9 people.
Thus ways to make groups of 3 people groups have atleast 1 senior.
= 4C1*9C2
hope this clarifies.

Kudos [?]: 13 [0], given: 0

CEO
Joined: 15 Aug 2003
Posts: 3454

Kudos [?]: 919 [0], given: 781

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23 Sep 2003, 09:55
Vicky wrote:
oh man.. silly mistake...
thanks for correction.
Second solution:
4C1 = number of ways one senior can be selcted
9C2 = number of ways any 2 people can be selected from remaining 9 people.
Thus ways to make groups of 3 people groups have atleast 1 senior.
= 4C1*9C2
hope this clarifies.

1. C ...both are required to solve the problem

Most simplest way is 10C3 - 6C3

Vicks...did you notice the ATLEAST one senior part in the problem

Question, why do you multiply the combinations?

Kudos [?]: 919 [0], given: 781

SVP
Joined: 30 Oct 2003
Posts: 1788

Kudos [?]: 112 [0], given: 0

Location: NewJersey USA

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16 Jan 2004, 14:24
1+X+X^2+X^3+X^4+X^5 < 1/(1-X) ?

From a and b we can say x is between 0 and 1
if x = 0.9 then 1+X+X^2+X^3+X^4+X^5 < 1/(1-X)
if x = 0.5 then 1+X+X^2+X^3+X^4+X^5 > 1/(1-X)

how can we determine this for sure ?

Kudos [?]: 112 [0], given: 0

SVP
Joined: 30 Oct 2003
Posts: 1788

Kudos [?]: 112 [0], given: 0

Location: NewJersey USA

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16 Jan 2004, 14:27
I did a mistake. for x = 0.5

1+X+X^2+X^3+X^4+X^5 < 1/(1-X)

I take back my post.

Limit of N -> Infinity for 0<x<1
1 + Summation( X ^ N ) -> 1/(1-X)

Kudos [?]: 112 [0], given: 0

16 Jan 2004, 14:27
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