utin wrote:

There are 8 students. 4 of them are men and 4 of them are women. If 4 students are selected from the 8 students. What is the probability that the number of men is equal to that of women?

A. 18/35

B. 16/35

C. 14/35

D. 13/35

E. 12/35

According to me answer shall be= 4/8 * 3/7 * 4/6 * 3/5

Were I am wrong???Please explain

METHOD-1:Favorable outcomes (i.e. No. of Men= No. of Women =2) = 4

C2 * 4

C2 = 6*6 = 36

Total ways of selecting 4 out of 8 Students = 8

C4 = 8! / (4! * 4!) = 70

Probability = 36/70 = 18/35

Answer: Option

METHOD-2: Also see the mistake done by the person who posted this questionProbability of First selected person being Man = 4/8

Probability of Second selected person being Man = 3/7

Probability of First selected person being Woman = 4/6

Probability of Second selected person being Woman = 3/5

i.e. Probability = (4/8)*(3/7)*(4/6)*(3/5) *

[4!/(2!*2!)] = 18/35

The important part for the readers is to understand the reason of multiplying [4!/(2!*2!)] hereWhen we take the probability of each case like we have have taken in this method then it always include ARRANGEMENTS as well and so we have to take every arrangement of the events as well

The arrangement of these 4 events can be done in 4! ways but since the second man can't be selected before 1st so we have to exclude their arrangement by dividing by 2! and similarly since the second Woman can't be selected before 1st so we have to exclude their arrangement by dividing by 2!Answer: Option

_________________

Prosper!!!

GMATinsight

Bhoopendra Singh and Dr.Sushma Jha

e-mail: info@GMATinsight.com

Call us : +91-9999687183 / 9891333772

http://www.GMATinsight.com/testimonials.html

Contact for One-on-One LIVE Online (SKYPE Based) Quant/Verbal FREE Demo Class

______________________________________________________

Please press the

if you appreciate this post !!