- A student is to answer 10 out of 15 questions in an examination such that he must choose at least 5 from the first six questions. The number of choices available to him is
No of choices=6c5.9c5 +6c6.9c4
=6*[(9*8*7*6)/(4*3*2*1)] + 1*[(9*8*7*6)/(4*3*2*1)] =6*126 +1*126
- The number of ways in which a team of ten players can be selected from 20 players including 2 of them and excluding 4 of them is
2 players already included.
Now 8 players have to selected from 14 players.
- In how many ways 3 prizes can be given away to 10 boys when each boy is eligible for any of the prizes?
No of ways=10*10*10=1000ways.
- 3 boys and 2 girls are to be seated in a row in such a way that two girls are always together. In how many different ways can they be seated?
2Girls always together then consider them as one.
Now there are 4 members.
And that two girls can sit in 2! Ways
Then No of ways=4!*2!
- How many four digit number can be formed whose unit digits is always zero and repetition of the digit is not allowed?
In a 4 digit no unit digit is fixed ie 0.
Then in remaining 3 digit is from no 1-9
Then no of ways=9*8*7*1= 504.
- From 6 men and 3 women, a committee of 5 member is to be formed. In how many ways, if the committee has to include at least one women?
Ways= (1 women *4men) +(2 women*3men)+ (3 women *2men)
=3c1*6c4 +3c2*6c3 + 3c3*6c2
=3*15 + 3*20 +1*15
- In how many ways can a leap year have 53 Sundays?
In a leap year there are 366 days i.e. 52 weeks + 2 extra days.
Then 53 Sundays can occur in only 2 ways.
i.e. (Saturday and Sunday) or (Sunday and Monday).
no of ways =2.
- There are five trains running between station A and B. In how many ways can a man go from A to B and return by a different train?
A Man go from A to B in 5 ways by 5 train available.
He return from B to A in 4 ways by the remaining 4 train only.
Then no of ways=5*4=20ways.
- How many words can be formed by re-arranging the letters of the word CURRENT such that C and T occupy the first and last position respectively?
First and last position stable.
C_ _ _ _ _T
Remaining URREN must be change
- How many different straight lines can be formed by joining 12 different points on a plane of which 4 are collinear and the rest are non-collinear?
Total no of lines formed by 12 points=12c2
No of lines forms by 4 collinear points=4c2
Then Required no of lines=12c2 – 4c2 +1