Quantitative aptitude question with answer permutation and combination solved

In  campus interview quantitative aptitude question and  answer permutation and combination  are important so we provide many question and answer here:

Combinations and permutation:shortcuts

The number of different groups of r objects that can be formed from a total of n objects when the order of selection is irrelevant is called the number of combinations of n objects taken r at a time.  
The number of these r combinations is denoted by ⁿC_rorC(n, r)and

Recall that ⁿP_r is the number of ways in which r objects can be selected from n objects, when the order of selections is relevant.
We can also choose an ordered selection of r object out of n objects in two steps.
1. Choose a subset consisting of r objects without paying any attention to the orderin which they are chosen. This can be done in ⁿC_r ways.
2. For a given choice of r objects, count the number of ways in which these object could be ordered. Any r objects can be permuted amongst them in r! ways.

Hence from the fundamental rule of counting we must have :

The second and the third equalities follows from the fact that:

(a) In a game of poker a player receives 5 cards from an ordinary deck of playing cards. These 5 cards are called a poker hand. How many poker hands are there?    
(b) A poker hand is said to be a flush if all the cards are of the same suit. In how many ways can one be dealt a flush?
Solution:

(a) The number of poker hands that a player can receive is

(b) There are four suits in a deck of cards. Out of these 4 suits, one suit has to be chosen and from this suit 5 cards have to be chosen (out of 13 cards in the four suits). The number of ways one can be dealt a flush is:

Example:

(a) A student has to answer 6 out of 10 questions in an examination. How many choices does she have?
(b) How many choices are there if she must answer at least 2 of the first 3 questions?

Solution:

(a) The student has to select 6 out of 10 of questions. There are

choices for her.

(b) If the student must answer at least 2 of the first 3 questions, it is convenient to divide her choices into tow mutually exclusive groups.
(i) She answers all three of the first three questions and answers any 3 of the remaining 7 questions. This can be done in

(ii) She chooses 2 out of the first 3 questions and answers any 3 of the remaining 7 questions. This can be done in

Hence the number of choices that are available to the student is

35 + 105 = 140

Example:

A man has 10 friends whom he wants to invite for dinner. The number of ways in which he can invite at least one of them is:
[1] 1024

[2] 10!

[3] 1023

[4] 10! - 1
Solution:
Suppose the man has to invite only one of his friends (out of 10) for dinner, then we know that he has to choose one amongst his 10 friends. This can be done in: 10_C1ways
The key word in the question is: at least, this means the friend can call 1 or 2 or 3 or…10 friends over for dinner. Mathematically, this can be shown as:

Therefore, the man can invite his 10 friends in (1024 – 1) or 1023 ways.
Example:
Assume the above problem is framed as follows:
A man has 10 friends whom he wants to invite for dinner. The number of ways in which he can invite at least three of them is:
(1) 1021                        

(2) 968

(3) 864                         

(4) None of these
Solution:

Suppose the man has to invite only three of his friends (out of 10) for dinner, then we know that he has to choose three amongst his 10 friends. This can be done in: 10_C3
The key word in the question still is: at least, this means the friend can call 3 or 4 or 5 or…10 friends over for dinner. Mathematically, this can be shown as:

Therefore, the man can invite his 10 friends in [(1024 – 1)– (10+45)]or 968 ways.
Example:
A question paper has 5 multiple choice questions and 5 true or false questions. Each multiple choice question is provided with 3 choices. Find the number of ways that an examinee can attempt one or more questions of the given 10 questions.
(1)  3¹⁰ (2) 6⁵-1
(3)  8⁵ -1    (4) 12⁵ -1

Solution:

Number of ways of dealing with each multiple choice question = 4(3 ways of attempting and one way of not attempting)
Number of ways of dealing with each true or false question = 3 (2 ways of attempting and one way of not attempting)
Therefore Number of ways of attempting one or more questions = (4⁵ × 3⁵) – 1 = 12⁵ – 1