# From a deck of 52 cards, a 5 card hand is dealt.How may distinct five card hands are there if the queen of spades and the four of diamonds must be in the hand?

 A) 52C5 B) 50C3 C) 52C4 D) 50C4
Explanation:

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, $\inline&space;{\color{Blue}&space;50C_{3}}$

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### A local delivery company has three packages to deliver to three different homes. if the packages are delivered at random to the three houses, how many ways are there for at least one house to get the wrong package?

 A) 3 B) 5 C) 3! D) 5!

Explanation:

The possible outcomes that satisfy the condition of "at least one house gets the wrong package" are:
One house gets the wrong package or two houses get the wrong package or three houses get the wrong package.

We can calculate each of these cases and then add them together, or approach this problem from a different angle.
The only case which is left out of the condition is the case where no wrong packages are delivered.

If we determine the total number of ways the three packages can be delivered and then subtract the one case from it, the remainder will be the three cases above.

There is only one way for no wrong packages delivered to occur. This is the same as everyone gets the right package.

The first person must get the correct package and the second person must get the correct package and the third person must get the correct package.
1×1×1=1

Determine the total number of ways the three packages can be delivered.
3×2×1=6

The number of ways at least one house gets the wrong package is:
6−1=5
Therefore there are 5 ways for at least one house to get the wrong package.

Subject: Permutation and Combination - Quantitative Aptitude - Arithmetic Ability

### Jay wants to buy a total of 100 plants using exactly a sum of Rs 1000. He can buy Rose plants at Rs 20 per plant or marigold or Sun flower plants at Rs 5 and Re 1 per plant respectively. If he has to buy at least one of each plant and cannot buy any other type of plants, then in how many distinct ways can Jay make his purchase?

 A) 3 B) 6 C) 4 D) 2

Explanation:

Let the number of Rose plants be ‘a’.
Let number of marigold plants be ‘b’.
Let the number of Sunflower plants be ‘c’.
20a+5b+1c=1000; a+b+c=100

Solving the above two equations by eliminating c,
19a+4b=900

$\inline&space;{\color{Blue}b=\frac{900-19a}{4}}$

$\inline&space;{\color{Blue}b=225-\frac{19a}{4}}$----------(1)

b being the number of plants, is a positive integer, and is less than 99, as each of the other two types have at least one plant in the combination i.e .:0 < b < 99--------(2)

Substituting (1) in (2),

0 < $\inline&space;{\color{Blue}225-\frac{19a}{4}}$ < 99

$\inline&space;{\color{Blue}225<-\frac{19a}{4}<99-225}$

$\inline&space;{\color{Blue}\Rightarrow&space;4\times&space;225>19a>126\times&space;4}$

$\inline&space;{\color{Blue}\Rightarrow&space;\frac{900}{19}>a>504}$
a is the integer between 47 and 27 ----------(3)
From (1), it is clear, a should be multiple of 4.

hence possible values of a are (28,32,36,40,44)

For a=28 and 32, a+b>100
For all other values of a, we get the desired solution:
a=36,b=54,c=10
a=40,b=35,c=25
a=44,b=16,c=40

Three solutions are possible.

Subject: Permutation and Combination - Quantitative Aptitude - Arithmetic Ability

### From a total of six men and four ladies a committee of three is to be formed. If Mrs. X is not willing to join the committee in which Mr. Y is a member, whereas Mr.Y is willing to join the committee only if Mrs Z is included, how many such committee are possible?

 A) 91 B) 104 C) 109 D) 98

Explanation:

We first count the number of committee in which

(i). Mr. Y is a member
(ii). the ones in which he is not

case (i): As Mr. Y agrees to be in committee only where Mrs. Z is a member.
Now we are left with (6-1) men and (4-2) ladies (Mrs. X is not willing to join).

We can choose 1 more in5+$\inline&space;{\color{Blue}&space;2C_{1}}$=7 ways.

case (ii): If Mr. Y is not a member then we left with (6+4-1) people.
we can select 3 from 9 in $\inline&space;{\color{Blue}&space;9C_{3}}$=84 ways.

Thus, total number of ways is 7+84= 91 ways.

Subject: Permutation and Combination - Quantitative Aptitude - Arithmetic Ability

### There are five cards lying on the table in one row. Five numbers from among 1 to 100 have to be written on them, one number per card, such that the difference between the numbers on any two adjacent cards is not divisible by 4. The remainder when each of the 5 numbers is divided by 4 is written down on another card (the 6th card) in order. How many sequences can be written down on the 6th card ?

 A) 4 x 3^4 B) 3^4 C) 4^3 D) 3 x 4^3
Answer & Explanation Answer: A) 4 x 3^4

Explanation:

The remainder on the first card can be 0,1,2 or 3 i.e 4 possibilities.
The remainder of the number on the next card when divided by 4 can have 3 possible values (except the one occurred earlier).

For each value on the card the remainder can have 3 possible values.

The total number of possible sequences is: $\inline&space;{\color{Blue}4\times&space;3^{4}}$

Subject: Permutation and Combination - Quantitative Aptitude - Arithmetic Ability

### There are eight boxes of chocolates, each box containing distinct number of chocolates from 1 to 8. In how many ways four of these boxes can be given to four persons (one boxes to each) such that the first person gets more chocolates than each of the three, the second person gets more chocolates than the third as well as the fourth persons and the third person gets more chocolates than fourth person?

 A) 70 B) 40 C) 72 D) 80
The number of ways of giving 4 boxes to the 4 person is: $\inline&space;{\color{Blue}8C_{4}}$= 70