Questions about the world of GMAT Math from other sources and general math related questions.
riyakhilnani7
Forum Guests
 
Posts: 3
Joined: Mon May 28, 2012 3:35 am
 

Tough permutations/combinations questions

by riyakhilnani7 Sat Sep 22, 2012 5:46 am

Hey, can anyone suggest ways to solve these problems?

Thanks!

1) How many 10 digit numbers can be formed using 3 and 7 only?

2) How many 5 digit numbers can be formed which are divisible by 3 using the numerals 0,1,2,3,4,5 (without repitition)

3) Everyone shakes hands with everyone else in a room. Total number of handshakes is 66. Number of persons=?

4) The number of parallelograms that can be formed from a set of FOUR parallel straight line intersecting a set of THREE parallel straight lines=?
garima_aries01
Students
 
Posts: 9
Joined: Tue Jun 29, 2010 8:52 pm
 

Re: Tough permutations/combinations questions

by garima_aries01 Sat Sep 22, 2012 7:19 am

riyakhilnani7 wrote:1) How many 10 digit numbers can be formed using 3 and 7 only?


10 digit no , each digit can can be either 3 or 7 => each digit can be occupied by 2 numbers
_ _ _ _ _ _ _ _ _ _
2 2 2 2 2 2 2 2 2 2

=> 2^10 numbers

riyakhilnani7 wrote:2) How many 5 digit numbers can be formed which are divisible by 3 using the numerals 0,1,2,3,4,5 (without repitition)


for the number to be divisible by 3 , the sum of its digits should be divisble by 3
0+1+2+3+4+5 = 15

therefore the digits in the number can be either
01245 (sum of its digits = 12)
or
12345 (sum of its digits = 15)

If digits are 12345, there can be 5!=120 numbers without repetition of the digits
If digits are 01245, there can be 4x4x3x2x1 = 96 numbers without repetition of the digits
because _ _ _ _ _
4 4 3 2 1
the 1st digit cant be zero otherwise it will be 4 digit no.
total 120+96 = 216 numbers

riyakhilnani7 wrote:3) Everyone shakes hands with everyone else in a room. Total number of handshakes is 66. Number of persons=?


Suppose n people are there. When n people are taken 2 at a time there are 66 handshakes so
nC2 = 66
n!/[2!x(n-2)!] = 66
=> n = 12

riyakhilnani7 wrote:4) The number of parallelograms that can be formed from a set of FOUR parallel straight line intersecting a set of THREE parallel straight lines=?


We can assume there are 4 horizontal parallel lines ,3 vertical parallel lines
parallelogram is made by 2 horizontal parallel lines and 2 vertical parallel liness
so
4C2 x 3C2 = 18 parallelograms
riyakhilnani7
Forum Guests
 
Posts: 3
Joined: Mon May 28, 2012 3:35 am
 

Re: Tough permutations/combinations questions

by riyakhilnani7 Sat Sep 22, 2012 7:48 am

thanks for the tips. one more... A polygon has 44 diagonals. Number of sides=?

A. 7
B. 11
C. 8
D. 9
E. 10
garima_aries01
Students
 
Posts: 9
Joined: Tue Jun 29, 2010 8:52 pm
 

Re: Tough permutations/combinations questions

by garima_aries01 Sat Sep 22, 2012 11:55 am

riyakhilnani7 wrote: A polygon has 44 diagonals. Number of sides=?


Suppose polygon has n sides
diagnol is formed by joining 2 points
nC2
BUT we need to subtract n from nC2 to exclude to sides of the polygon
So,
nC2-n = 44
n^2-3n-88=0
n=11 OR n=-8

=> n=11 as n cant be negative
tim
ManhattanGMAT Staff
 
Posts: 5324
Joined: Tue Sep 11, 2007 9:08 am
Location: Southwest Airlines, seat 21C
 

Re: Tough permutations/combinations questions

by tim Wed Sep 26, 2012 11:41 am

thanks! let us know if there are any further questions on these..
Tim Sanders
Manhattan GMAT Instructor

Follow this link for some important tips to get the most out of your forum experience:
https://www.manhattanprep.com/gmat/forums/a-few-tips-t31405.html