Permutations and Combinations Test - 5

Given that there are 20 stations in the network .

Hence from each station 19 different tickets are possible

Therefore   from 20 stations the number of  different tickets  possible=19 X 20=380

First we will fix one person  from  the 6 boys  then  5 others can be  arranged in  5! ways=120 ways 

Now there are 6 places left in which 2 brothers can sit,so they can choose any 2 places from  the 6 places  in 6C2 ways=15 ways

Also 2 brothers can arrange themselves in 2! ways

So th ways in which the two brothers can be seated=15*2=30

Hence total ways in which all can be seated =120*30=3600

Since a student can solve every question in three ways- either he can attempt the first alternative , or the second alternative  or he does not attemp that question 

Hence the total ways in which a sudent can attempt one or more of  8 questions =3⁸

Therefore  to find the number of all selections which a student can make for answering one or more questions ou tof 8 given questions =3⁸−1 = 6560   _{[ we will have to exclude  only the case of not answering all the 8 questions]}

4 flowers which are always together can be considered as one SET, 

Therefore we have to arrange one SET ( 4 flowers ) and 4 other flowers into a garland. 

Which means, 5 things to be arranged in a garland. 

(5-1)! 

And the SET of flowers can arrange themselves within each other in 4! ways. 

Therefore 

(5-1)!*(4!) 

But, Garland, looked from front or behind does not matter. Therefore the clockwise and anti clockwise observation does not make difference. 

Therefore 

(5-1)! * (4!)/ 2 

= 288.

The month of February  has either 28 days (non-Leap Year)  or 29 days(Leap Year).

Thus there are 2 possibility  for the number of days in February.

Also the first day of the month of February can be any of the 7 days in a week.

So total number of February calendars = 2 * 7 =14

we have 3600=2⁴.3².5²

To get the odd factors we will get rid of 2's

We will make the selection from only 3's and 5's 

Number of ways 3 can be selected from a lot of two 3's= 3 ways ( one 3,two 3's or three 3's)

Number of ways 5 can be selected from a lot of two 5's= 3 ways ( one 5,two 5's or three 5's)

Therefore  the number of odd factors is 3600= 3 X 3 =9

We have 1600=2⁶.5²

To form factors we have to do selections from a lot of 2's and 5's and multiply them together.

To form even factor we should choose at least one 2's from the lot , which will ensure that what ever be the remaining selection, their multiplication will always result in an even factor.

The number of ways to select atleast one ‘2’ from a lot of six  identical ‘2’s will be 6 (i.e. select 1 or select 2 or select 3 or select 4 or select 5 or select 6)

And, we’ll select any number of ‘5 from a lot of two   identical ‘5’s in 3 ways(select 0, select 1,select 2)

There fore the total number of selection of even factors=6x3=18
 

Given xyz=30

We have the possible values of x ,y,z are the following triads 

1,1,30 
1,2,15 
1,3,10 
1,5,6 
2,3,5 
First one can have 3!/2! = 3 ways and the remaining four triads can have 3! combinations 
Hence total combinations = 3 + 4*3! = 27