Challenge Problem this week

[elawyer8]

Dear Manhattan GMAT:

There appears to be a typo error in this week's challenge problem (currently displayed on the Home page of this site). The answer choice C: (7! - 5!3!) is "visually" closest to the answer that I think is correct (7! - (5 x 3!)) , but not quite correct, I think.

Regards,

[cindyqtran]

wow, i thought it was either A or B. not sure. I am not sure if the position of the women matter here.

How did u get (5!) in answer choice C?

[shaji]

C is the correct answer. GMAT sometimes not too good at extrasensory deceptions. Visually close answers are often correct, hence the 'challenge'.

Dear Manhattan GMAT:

There appears to be a typo error in this week's challenge problem (currently displayed on the Home page of this site). The answer choice C: (7! - 5!3!) is "visually" closest to the answer that I think is correct (7! - (5 x 3!)) , but not quite correct, I think.

Regards,

[StaceyKoprince]

Answer choice C is presented correctly: (7! - 5!3!). I won't go into whether this is the right answer right now, since this question is still "open" - see if you can figure it out and check back next week when the answer is posted. :)

[shaji]

A quick fix for GMAT problems !!!

Having elliminated the impossible, then whatever is left, however improbable it may seem, has to be the correct answer!!!

Since the problem is still open, I shall post a complete mathematical analysis of this problem exhausting all possibilities next Monday.

C is the correct answer. GMAT sometimes not too good at extrasensory deceptions. Visually close answers are often correct, hence the 'challenge'.

Dear Manhattan GMAT:

There appears to be a typo error in this week's challenge problem (currently displayed on the Home page of this site). The answer choice C: (7! - 5!3!) is "visually" closest to the answer that I think is correct (7! - (5 x 3!)) , but not quite correct, I think.

Regards,

[GMAT 2007]

The answer is (C) and it is stated correctly too. Here is the solution: -

We are looking for the cases when three men will not sit in the three adjacent seats: -

Desired cases when men will not sit adjacent to each other = Total no. of ways of seating arrangement - No. of ways men will sit at the three adjacent seats

Total no. of ways men & women can be seated= 7!

Now lets go through the possible ways men can be seated in the three adjacent seats: -

1 2 3 4 5 6 7 No of Ways
M1 M2 M3 W W W W = 4!3!
W M1 M2 M3 W W W = 4!3!
W W M1 M2 M3 W W = 4!3!
W W W M1 M2 M3 W = 4!3!
W W W W M1 M2 M3 = 4!3!

So desired cases = 7! - 5X4!3! = 7! -5!3!

So the correct answer is (C)

Hope it helps

GMAT 2007

[SHAJI]

Precisely!!! C is the correct answer as stated and it was never in doubt. However, a complete audit of all 7! possibilities from the man's point of view and and the women's point of view independantly would be an interesting exercise. Dealing with similar situations on the GMAT would then be a 30 second job.

The answer is (C) and it is stated correctly too. Here is the solution: -

We are looking for the cases when three men will not sit in the three adjacent seats: -

Desired cases when men will not sit adjacent to each other = Total no. of ways of seating arrangement - No. of ways men will sit at the three adjacent seats

Total no. of ways men & women can be seated= 7!

Now lets go through the possible ways men can be seated in the three adjacent seats: -

1 2 3 4 5 6 7 No of Ways
M1 M2 M3 W W W W = 4!3!
W M1 M2 M3 W W W = 4!3!
W W M1 M2 M3 W W = 4!3!
W W W M1 M2 M3 W = 4!3!
W W W W M1 M2 M3 = 4!3!

So desired cases = 7! - 5X4!3! = 7! -5!3!

So the correct answer is (C)

Hope it helps

GMAT 2007

[dbernst]

Just a little word to the wise about the challenge problems. The challenge problems are great fun, and they are also excellent practice for those scoring in the 95th+ percentile on their Quant sections. However, most of the challenge problems are somewhere between 750 and 900 in terms of difficulty (that is not a typo - many of the challenge problems are more difficult than any problem you will see on your official GMAT). Thus, though I encourage you to continue to work through the challenge problems, do not do so at the expense of truly solidifying your fundamentals.

Unlike Stacey K and Emily S and Jad L, who score in the 99th percentile on their Quant, I only score in the 90-93 percentile range (I like to think that I return the favor on the verbal, but they all score exceptionally on it as well!) Thus, I rarely see problems on my quant sections that are as difficult as the challenge problems. Because I am aware of the types of problems I am most likely to see (at least in terms of difficulty), I prepare for my GMAT accordingly. Whereas my colleagues might do challenge problems galore, I rarely look at them. For me, it's all about the fundamentals (and I mean FUNdamentals!).

Just my 2 cents. Hope it helps!
-dan

[GMAT 2007]

Thanks for your 2 cents :) Dan

[shaji]

Its Monday, and the correct answer is posted for all to see. Now the rest of the story!!!

# of ways that each of the three men are flanked by women on both sides=144
#of ways that two women are together while the other two are seperated by men=1728
#of ways that the women sit in pairs seperated by men=2592
# of ways the all four women are together =576

This is the complete audit of the 7! ways.

Precisely!!! C is the correct answer as stated and it was never in doubt. However, a complete audit of all 7! possibilities from the man's point of view and and the women's point of view independantly would be an interesting exercise. Dealing with similar situations on the GMAT would then be a 30 second job.

The answer is (C) and it is stated correctly too. Here is the solution: -

We are looking for the cases when three men will not sit in the three adjacent seats: -

Desired cases when men will not sit adjacent to each other = Total no. of ways of seating arrangement - No. of ways men will sit at the three adjacent seats

Total no. of ways men & women can be seated= 7!

Now lets go through the possible ways men can be seated in the three adjacent seats: -

1 2 3 4 5 6 7 No of Ways
M1 M2 M3 W W W W = 4!3!
W M1 M2 M3 W W W = 4!3!
W W M1 M2 M3 W W = 4!3!
W W W M1 M2 M3 W = 4!3!
W W W W M1 M2 M3 = 4!3!

So desired cases = 7! - 5X4!3! = 7! -5!3!

So the correct answer is (C)

Hope it helps

GMAT 2007

[Rustic Myth]

Hi I do not know the question but by going through the discussion the simple way would be.

MMMWWWW ----------- 7 people

total no of ways =7P7 = 7!

As three men should be together MMM = X

the can be seated as 3P3 = 3!

XWWWW ----------- 5 Seats considering MMM = X

so total no of ways = 5P5 = 5!

so the number of ways possible = 7! - 5!3!

:)

i think some time grouping similar things helps in probablity......... !!! :oops: :oops: