Community

Want to learn how to tackle those tough AMC/AIME/Olympiad counting and probability problems? Check out Art of Problem Solving's Intermediate Counting & Probability by David Patrick.
Login Register Memberlist Search AoPS Blogs Contests Galleries Forum Index
The time now is Thu Dec 03, 2009 1:09 am
All times are UTC - 8
 		View posts since last visit
View unanswered posts
Home » Mathematics » High School Intermediate Topics (Ages 13+)
 
combinatoric
Moderators: Intermediate Topics Moderators
Post new topic   Reply to topic View previous topicView next topic
3 Posts • Page 1 of 1
Author Message
zerensabri
New Member
New Member

Offline
Joined: 03 Nov 2009
Posts: 1
To rate posts you must be logged in
#1
combinatoric
writing different numbers
1) How many different six digit numbers are there whose three digits are even and three digits are odd ?
2) How many different 4-digit numbers can be written by using two digits-one is prime and the other is not- chosen from the set A= {1,2,3,4,5,6,7,8,9} ?

PostPosted: Sat Nov 07, 2009 7:47 am  Back to top 
  ProfilePM
grn_trtle
Navier-Stokes Equations
Navier-Stokes Equations


Offline
Joined: 01 Sep 2008
Posts: 1016
Location: California
IndiaUnited States
To rate posts you must be logged in
#2
Well...

1

\binom{6}{3} ways to pick the even digits (or odd ones)
5^3 ways to choose three even digits.
5^3 ways to choose three odd digits.
\binom{6}{3}(5^3)^2 = 312500.
Out of these, \binom{5}{3}5^35^2=31250 start with 0, making the answer \boxed{281250}.


2

There are 4 primes: 2,3,5,7
4\cdot 5 = 20 ways to pick one prime and one non-prime
Then \binom{4}{2} ways to pick two digits to be the prime ones.
So, \boxed{120}.


Edit: Yeah, tonypr is right, fixed it
_________________
\arctan\alpha + \arctan\beta = \text{arg}\{(1+\alpha i)(1+\beta i)\} = \arctan\left( \frac{\alpha+\beta}{1-\alpha\beta} \righ...
Last edited by grn_trtle on Sat Nov 07, 2009 9:52 pm; edited 1 time in total 
PostPosted: Sat Nov 07, 2009 1:56 pm  Back to top 
  ProfilePM
tonypr
Riemann Hypothesis
Riemann Hypothesis


Offline
Joined: 25 Oct 2007
Posts: 485
Location: San Juan, Puerto Rico
Puerto Rico
To rate posts you must be logged in
#3
grn_trtle wrote:
Well...
\binom{6}{3} ways to pick the even digits (or odd ones)
5^3 ways to choose three even digits.
5^3 ways to choose three odd digits.
\binom{6}{3}(5^3)^2 = \boxed{312500}.


Don't you have to discount the ones where 0 is the first digit?
_________________
Math in trivial problem solving

PostPosted: Sat Nov 07, 2009 2:33 pm  Back to top 
  ProfilePMBlog
Display posts from previous:   Sort by:   
3 Posts • Page 1 of 1
Post new topic   Reply to topic View previous topicView next topic
Home » Mathematics » High School Intermediate Topics (Ages 13+)
 
Jump to:  


You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot vote in polls in this forum
You cannot attach files in this forum
You can download files in this forum
You cannot post calendar events in this forum


© Copyright 2008 AoPS Incorporated. All Rights Reserved. • Foundation • Privacy • Contact Us