GRE Question of the Day: Permutation and Combination Comparison

avinash
#1. June 25th, 2006, at 2:06 AM.

The answer is B)….
the total number of possible ways of inviting atleast one person is…..as follows..
10c1+10c2+10c3+……..+10c10

the abv expression is equal to 2^10 -10c0=1024-1=1023…
hence option B)..

Dale
#2. June 25th, 2006, at 3:11 AM.

How come you guys are so good at probobilities????
Wowwww

avinash
#3. June 25th, 2006, at 3:30 PM.

just brush ur basics m8 ..understandin the copncept is damn important in permu n comb…the more u understan the concept the more ull b intrested in solvin such type of probs..n the more u do the probs the more ull understand the concepts..:)

abhi.roy
#5. June 15th, 2007, at 10:35 AM.

B

Reynard
#6. June 15th, 2007, at 1:43 PM.

Answer is B
Beside Avinash’s explanation, there’s still another way to reach the correct answer.

Imagine you have a set of 10 friends. So there are 2^10 = 1024 subsets of that, including the empty set. So the number of ways to invite at least 1 friend is the number of subsets, excluding the empty set (which implies inviting none of the friends): 2^10 - 1 = 1023

Ailiam Attobra
#7. June 22nd, 2007, at 2:32 AM.

B

vidya
#8. August 23rd, 2007, at 4:34 PM.

b

gops
#9. August 23rd, 2007, at 5:06 PM.

B

benign
#10. August 23rd, 2007, at 9:09 PM.

b

anushree
#11. August 24th, 2007, at 2:29 PM.

ans [b]

AKII
#12. August 26th, 2007, at 8:30 PM.

b

varunsingh
#13. August 27th, 2007, at 10:14 AM.

(b)

anj
#14. August 28th, 2007, at 11:04 AM.

where does the 2^10 come into the pic?

asifiqbal
#15. August 28th, 2007, at 12:44 PM.

it’s B

minal
#16. August 29th, 2007, at 7:58 PM.

B
becoz it can be calculated as 10*10……10times

sandeep9
#17. September 4th, 2007, at 3:27 PM.

(2^10)-1=1023
so answer is B

harish
#18. September 16th, 2007, at 1:22 PM.

b

Hitesh
#19. September 20th, 2007, at 12:33 PM.

B

Vineet Pandit
#20. October 19th, 2007, at 4:01 PM.

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

nC0 + nC1 + nC2 + . . . + nCn = 2n.
THERE FORE

nC1 + nC2 + . . . + nCn = 2n - nC0

prash_xp
#21. November 9th, 2007, at 3:13 AM.

Sure the answer is B) since …. 1024 > 1023.

I think its pretty straightforward question…and many beautiful approaches have been listed already, so I am in a dilemma whether to discuss my approach also ..

……… having said that, cannot resist giving my way of arriving at the answer..kindly bear

For each of his 10 friends, he has 2 choices — whether to invite or not.
So total options, 2 x 2 x 2 …. (10 times) = 1024.

But this figure of 1024 has that option also that says “Dont invite” for all of his 10 friends..

We are required to find the number of ways to invite “at least” 1 friend..so we need to reject that option, giving us 1024 - 1 = 1023 options

naresh
#22. January 30th, 2008, at 4:28 PM.

nc0+nc1+ +ncn= 2^n
10c1+ 10c2 + + 10c10 = 2^10- 10c0
= 1024 -1
col A = 1023
col B = 1024
Hence col B is greater
Ans -> (B)

mihir
#23. November 18th, 2008, at 10:38 PM.

yappy the answer is B.
hey guys is there any way to know the correct answer from a authorized source.
thanks

shilpa
#24. November 19th, 2008, at 11:38 AM.

yeah.. its B

Kofi
#25. November 19th, 2008, at 2:03 PM.

ok, so it’s B…

suhas r
#26. November 19th, 2008, at 5:58 PM.

b

rajesh
#27. November 20th, 2008, at 11:49 PM.

It’s B.

the explanation was same as what avinash gave…………..

sravanthi
#28. November 21st, 2008, at 9:34 AM.

the option both are equal
ans: 1+10c1+10c2+………+10c10=(1+1)^10=2^10=1024.

herzl
#29. November 22nd, 2008, at 4:08 AM.

B

simran
#30. December 9th, 2008, at 5:57 PM.

B