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# At a Pizza Parlor, in addition to cheese there are 10

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Director
Joined: 14 Jan 2007
Posts: 776
Followers: 2

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At a Pizza Parlor, in addition to cheese there are 10 [#permalink]

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23 Jun 2007, 06:05
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At a Pizza Parlor, in addition to cheese there are 10 different toppings. If you can order any number of toppings, then how many different toppings are possible.
VP
Joined: 08 Jun 2005
Posts: 1145
Followers: 7

Kudos [?]: 198 [0], given: 0

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23 Jun 2007, 07:12
C(10,0) = 1
C(10,1) = 10
C(10,2) = 45
C(10,3) = 120
C(10,4) = 210
C(10,5) = 252
C(10,6) = 210
C(10,7) = 120
C(10,8) = 45
C(10,9) = 10
C(10,10) = 1

total = 1,024

check = 2^10 = 1,024

Director
Joined: 14 Jan 2007
Posts: 776
Followers: 2

Kudos [?]: 141 [0], given: 0

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23 Jun 2007, 07:17
KillerSquirrel wrote:
C(10,0) = 1
C(10,1) = 10
C(10,2) = 45
C(10,3) = 120
C(10,4) = 210
C(10,5) = 252
C(10,6) = 210
C(10,7) = 120
C(10,8) = 45
C(10,9) = 10
C(10,10) = 1

total = 1,024

check = 2^10 = 1,024

That's correct.
Shortcut for such questions is -
nC0 + nC1 +nC2 +.........nCn = 2^n
Director
Joined: 26 Feb 2006
Posts: 903
Followers: 4

Kudos [?]: 114 [0], given: 0

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23 Jun 2007, 14:29
KillerSquirrel wrote:
C(10,0) = 1
C(10,1) = 10
C(10,2) = 45
C(10,3) = 120
C(10,4) = 210
C(10,5) = 252
C(10,6) = 210
C(10,7) = 120
C(10,8) = 45
C(10,9) = 10
C(10,10) = 1

total = 1,024

check = 2^10 = 1,024

interesting! I never noticed this one.

Is it always true for any n that 2^n?
VP
Joined: 08 Jun 2005
Posts: 1145
Followers: 7

Kudos [?]: 198 [0], given: 0

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23 Jun 2007, 14:54
Himalayan wrote:
KillerSquirrel wrote:
C(10,0) = 1
C(10,1) = 10
C(10,2) = 45
C(10,3) = 120
C(10,4) = 210
C(10,5) = 252
C(10,6) = 210
C(10,7) = 120
C(10,8) = 45
C(10,9) = 10
C(10,10) = 1

total = 1,024

check = 2^10 = 1,024

interesting! I never noticed this one.

Is it always true for any n that 2^n?

Yes ! "In general, the sum of all the combinations of n distinct things is 2^n.

nC0 + nC1 + nC2 + . . . + nCn = 2^n"

Last edited by KillerSquirrel on 24 Jun 2007, 14:27, edited 1 time in total.
Director
Joined: 26 Feb 2006
Posts: 903
Followers: 4

Kudos [?]: 114 [0], given: 0

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23 Jun 2007, 21:00
KillerSquirrel wrote:
Himalayan wrote:
KillerSquirrel wrote:
C(10,0) = 1
C(10,1) = 10
C(10,2) = 45
C(10,3) = 120
C(10,4) = 210
C(10,5) = 252
C(10,6) = 210
C(10,7) = 120
C(10,8) = 45
C(10,9) = 10
C(10,10) = 1

total = 1,024

check = 2^10 = 1,024

interesting! I never noticed this one.

Is it always true for any n that 2^n?

Yes ! "In general, the sum of all the combinations of n distinct things is 2n.

nC0 + nC1 + nC2 + . . . + nCn = 2^n"

grate learning. this concept directly goes to my note.

thanx..
23 Jun 2007, 21:00
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# At a Pizza Parlor, in addition to cheese there are 10

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