## Tough permutations/combinations questions

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

### Tough permutations/combinations questions

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
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Joined: Tue Jun 29, 2010 8:52 pm

### Re: Tough permutations/combinations questions

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
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Joined: Mon May 28, 2012 3:35 am

### Re: Tough permutations/combinations questions

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

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
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Joined: Tue Sep 11, 2007 9:08 am
Location: Southwest Airlines, seat 21C

### Re: Tough permutations/combinations questions

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