chetan2u wrote:
MathRevolution wrote:
How many possible cases are there such that 2 stocks are selected at random from 10 stocks?
A. 35
B. 40
C. 45
D. 50
E. 55
* A solution will be posted in two days.
Hi,
the Q is of COMBINATIONS but marked as PROBABILITY..
Please change it..
Method 1: No. of ways of selecting two out of 10 stocks = 10C2 = 10!/(2!*8!) = 45
Method 2: No. of ways of making pair of one stock with each of the remaining stock = 9
Total Such possible pairs (with Repetition) = 10*9 = 90
Since every case has been counted twice therefore total Unique pairs = 10*9 / 2 = 45
Method 3: Pairs of first stock with each of the 9 stocks can be made in 9 ways
Pairs of Second stock with each of the remaining 8 stocks can be made in 8 ways (second has already made pair
with first so we will not make that pair again
Pairs of Third stock with each of the remaining 7 stocks can be made in 7 ways (Third has already made pair with first and second so we will not make those pairs again) .... and so on...
Total Such pairs = 9+8+7+6+5+4+3+2+1 = 45
Answer: option C
_________________
Prosper!!!GMATinsightBhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhihttp://www.GMATinsight.com/testimonials.htmlACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION