Permutations and Combinations - Online Test

Since all the plus signs are identical ,  we have number of ways in which  6 plus signs  can be arranged=1.

Now we will have 7 empty slots between these 6 identical + signs 

Hence number of  possible places of - sign =7

Therefore number of ways in which the 4  minus sign  can take any of the possible 7 places=7C4=7!4!(7−4)!=5.6.71.2.3=35

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]

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

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

wehave3600=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
 

We have an n sided polygon has n vertices.

If you join every distinct pair of vertices you will get nC2 lines.

These nC2 lines account for the n sides of the polygon as well as for the diagonals.

So the number of diagonals is given by nC2–n=n(n−1)2–n=n(n−3)2

But given number of  sides = number of diagonals =n

⇒n=n(n−3)2⇒2n=n(n−3)⇒n−3=2⇒n=5