Robert Bosch Permutation and Combinations Quiz 1

first two places can be filled in 5 and 4 ways respectively so, total number of 3 digit number =
5*4*1 = 20

Vowels = E, E and A. They can be arranged in \frac{3!}{2!}
Ways so total ways = 6!*(\frac{3!}{2!}) = 2160

Zero or more bananas can be selected in 4 + 1 = 5 ways (0 orange, 1 orange, 2 orange, 3
orange and 4 orange)
similarly apples can be selected in 7 +1 = 8 ways
and mangoes in 6 +1 = 7 ways
so total number of ways = 5*8*7 = 280
but we included a case of 0 orange, 0 apple and 0 mangoes, so we have to subtract this, so 280
– 1 = 279 ways

From 15 points number of lines formed = ^{15}C_{2}
6 points are collinear, number of lines formed by these = ^{6}C_{2}
So total lines = ^{15}C_{2} –^{6}C_{2} + 1 = 91

4 Indians can be seated together in 4! Ways, similarly for Africans and Japanese in 5! and 7!
respectively. So total ways = 4! 5! 7! 3!

 

First Japanese can be seated along the circular table in 4! Ways and now Chinese can be
seated in 5! Ways. So 4! 5! Ways

Every chess match can have three result i.e. win, loss and draw
so now of ways = 3*3*3*3 = 81 ways

only 4 players to select, so it can be done in ^{10}C_{4} = 210

6 players to be selected from remaining 9 players in ^{9}C_{6} = 84 ways

^{6}C_{2}*^{5}C_{3} + ^{6}C_{3}*^{5}C_{2} + ^{6}C_{4}*^{5}C_{1}+ ^{6}C_{5}