no. of ways = 6!/3!2! = 60
why do we permute across 6 slots and account for the repeats? i dont understand the logic of it.
Thanks for making me rethink ....I thought on different lines...this is not the way this ques should be solved.
I think the correct approach is:
6C3 *3C2 *1C1 = 60
I think no. of ways = 6!/3!2! = 60 is correct.
6!=6P6 - all possible permutation with 3 wins, 2 losses and 1 draw
3!=3P3 - we exclude options for wins: (w1,w2,w3,l,l,d) and (w3,w1,w2,l,l,d) are the same way.
2!=2P2 - we exclude options for losses: (w,w,w,l1,l2,d) and (w,w,w,l2,l1,d) are the same way.
iOS/Android: GMAT ToolKit - The bestselling GMAT prep app | GMAT Club (free) | PrepGame | GRE ToolKit | LSAT ToolKit
PROMO: Are you an exiting GMAT ToolKit (iOS) user? Get GMAT ToolKit 2 (iOS) for free* (read more)
Math: GMAT Math Book ||| General: GMATTimer ||| Chicago Booth: Slide Presentation
The People Who Are Crazy Enough to Think They Can Change the World, Are the Ones Who Do.