umaa wrote:A certain panel is to be composed of exactly three women and exactly two men, chosen from x women and y men. How many different panels can be formed with these constraints?

(1) If two more women were available for selection, exactly 56 different groups of three women could be selected.

(2) x = y + 1

OA is C

(1) (w+2)C2 * yC2 = 56

(w+2)(w+1)/2! * y(y -1)/2! = 56

INSUFFICIENT

(2) INSUFFICIENT

(1) + (2) SUFFICIENT

aj5105, Thanks. But can you please explain it?

aj5105 wrote:

umaa wrote:A certain panel is to be composed of exactly three women and exactly two men, chosen from x women and y men. How many different panels can be formed with these constraints?

(1) If two more women were available for selection, exactly 56 different groups of three women could be selected.

(2) x = y + 1

OA is C

(1) (w+2)C2 * yC2 = 56

(w+2)(w+1)/2! * y(y -1)/2! = 56

INSUFFICIENT

(2) INSUFFICIENT

(1) + (2) SUFFICIENT

i think stmt 1 you have wrongly formulated...

we need to find xC3*yC2......

now as per stmt1,

if 2 more women were there then total women x+2

so the no. of ways selecting three women is (x+2)C3 and given that (x+2)C3=56

so x is known but still y not known. hence not sufficient.

but stmt 2 alone is also not sufficient because x& y both r unknown.

but if you use both stmt,

u may find x from first stmt first
and then y from second stmt.

hence C as xC3*yC2 can be calculated now.

The answer is C for sure, but I don't think the question can be correct...

There is no such value for x such that nCr(x, 3) = 56, so A doesn't make sense.

Domnu wrote:The answer is C for sure, but I don't think the question can be correct...

There is no such value for x such that nCr(x, 3) = 56, so A doesn't make sense.

no buddy! 8C3 surely makes 56. so there is a value of x available...

Ah, my mistake... I can't believe I did that