Quantitative Aptitude - Arithmetic Ability Questions

Two horses A and B, in a race of 6 horses... A has to finish before B

if A finishes 1... B could be in any of other 5 positions in 5 ways and other horses finish in 4! Ways, so total ways 5*4!

if A finishes 2... B could be in any of the last 4 positions in 4 ways. But the other positions could be filled in 4! ways, so the total ways 4*4!

if A finishes 3rd... B could be in any of last 3 positions in 3 ways, but the other positions could be filled in 4! ways, so total ways 3*4!

if A finishes 4th... B could be in any of last 2 positions in 2 ways, but the other positions could be filled in 4! ways, so total ways... 2 * 4! 

if A finishes 5th .. B has to be 6th and the top 4 positions could be filled in 4! ways..

A cannot finish 6th, since he has to be ahead of B

Therefore total number of ways = 5*4! + 4*4! + 3*4! + 2*4! + 4! = 120 + 96 + 72 + 48 + 24 = 360

There are 7 toppings in total,and by selecting 3,we will make different types of pizza.

This question is a combination since having a different order of toppings will not make a different pizza.

So,7C3 =35

since it is a combination = 38C17 = 2878 x 10^10

If John must be on the committee,we have 11 students remaining,out of which we choose 4. So,11C4

There are 13 spades,we must include 2: 13C2

There are 13 diamonds,we must include 3: 13C3

Since we can't have more than 2 spades and 3 diamonds,the remaining 2 cards must be pulled out from the 26 remaining clubs and hearts : 26C2

Therefore,13C2*13C3*26C2 = 7250100

If the queen of spades and the four of diamonds must be in hand,we have 50 cards remaining out of which we are choosing 3.

So, 50C3

Each candy can be dealt with in two ways.It can be chosen or not chosen.This will give 2 possibilites for the first candy, 2 for the second, and so on.by multiplying the cases together we get 212. Since the case of no candy being selected is not an option, we have to subtract 1 from our answer.

Therefore,there are 212- 1 = 4095 ways of selecting one or more candies

We get this using the formula of circular permutations i.e., (4 - 1)! = 3! =6