Permutations and Combinations MCQ Quiz - Objective Question with Answer for Permutations and Combinations - Download Free PDF

Calculation:

Given word = EXAMINATION

Total number of single letters = 8 (E, X, A, M, I, A, T, O)

Duplicate letters = 3 (AA, II, NN)

∴ Possible ways of selecting 4 letters are :

1.  4 from single = ⁸C₄ × 4! = 1680
2. 2 from single and 1 pair of the duplicated letters = ³C₁ × ⁷C₂ × \(\frac{4!}{2!}\) = 756
3. 2 pair of duplicated letters = ³C₂ × \(\frac{4!}{2!2!}\) = 18

Total number of ways = 2454

∴ The number of permutation of 4 letters taken from the word EXAMINATION is 2454.

The correct answer is Option 4.

Calculation

n-1Cr = (k2 - 8) nCr+1 

\(\underbrace{r+1 \geq 0, r \geq 0}_{r \geq 0}\)

⇒ \(\frac{{ }^{n-1} C_{r} }{{ }^nC_{r+1}}=k^2-8\)

⇒ \( \frac{r+1}{n}=k^2-8\)

⇒ k² - 8 > 0

⇒ \((k-2 \sqrt{2})(k+2 \sqrt{2})>0\)

⇒ \(k \in(-\infty,-2 \sqrt{2}) \cup(2 \sqrt{2}, \infty) \)                 --------(l)

\(\therefore n \geq r+1\Rightarrow\frac{r+1}{n} ≤ 1\)

⇒ k2​ - 8 ≤ 1 

⇒ k2 - 9 ≤ 0

⇒ -3 ≤ k ≤ 3                 --------(ll)

From equation (I) and (II) we get

\(\mathrm{k} \in[-3,-2 \sqrt{2}) \cup(2 \sqrt{2}, 3]\)

Hence option 1 is correct

Concept:

• The ways of arranging n different things = n!
• The ways of arranging n things, having r same things and rest all are different = \(\rm n!\over r!\)
• The no. of ways of arranging the n arranged thing and m arranged things together = n! × m!
• The number of ways for selecting r from a group of n (n > r) = nCr 

Calculation:

There are 3 vowels O, I, E

Required number of ways N = 1 vowel and 3 consonants + 2 vowel and 2 consonants + 3 vowel and 1 consonants

⇒ N = [\(\rm\left(^3C_1×{^3C_3}\right) + \left({^3C_2}×{^3C_2}\right) + \left({^3C_3}×{^3C_1}\right)\)] × 4!

⇒ N =  (3 + 9 + 3) × 4!

⇒ N = 15 × 4! words

Concept:

• The ways of arranging n different things = n!
• The ways of arranging n things, having r same things and rest all are different = \(\rm n!\over r!\)
• The no. of ways of arranging the n arranged thing and m arranged things together = n! × m!
• The number of ways for selecting r from a group of n (n > r) = nCr 
• To arrange n things in an order of a number of objects taken r things = nPr  

Calculation:

The operation can be done = (No token drawn is yellow) or (2 non-yellow token and 1 yellow token) or (1 non-yellow token and 2 yellow token)

Non-yellow tokens = 3 + 2 = 5

Ways of selecting of 3 non-yellow tokens out of 5 = 5C3 = 10

Ways of selecting of 2 non-yellow tokens out of 5 = 5C2 = 10

Ways of selecting of 1 non-yellow token out of 5 = 5C1 = 5 

Ways of selecting of 2 yellow tokens out of 5 = 5C2 = 10

Ways of selecting of 1 yellow token out of 5 = 5C1 = 5 

∴ The total number of ways N = 10 + (10 × 5) + (5 × 10)

N = 10 + 50 + 50 = 110

Formula used:

\(\displaystyle\sum_{n=0}^{k} \left(\! \begin{array}{c} n \\ k \end{array} \!\right) = 2^n\)

Calculation:

(7C0 + 7C1) + (7C1 + 7C2) + ...+(7C6 + 7C7)

⇒ 7C0 + 2.7C1 + 2.7C2 + ...+ 2.7C6 + 7C7 = x (say)   ....(1)

By using the above formula we have,

\(\displaystyle\sum_{n=0}^{k} \left(\! \begin{array}{c} n \\ k \end{array} \!\right) = 2^7\)

⇒ 7C0 + 7C1 + 7C2 + 7C3 + 7C4 + 7C5 + 7C6 + 7C7 = 27  ....(2)

Adding  7C0 and 7C7 on both sides of equation (1),we get

2.7C0 + 2.7C1 + 2.7C2 + ... + 2.7C6 + 2.7C7 = x + 7C0 + 7C7

⇒ 2 × 27 = x + 1 + 1

⇒ 28 = x + 2

⇒ 28 - 2 = x

⇒ 7C0 + 2.7C1 + 2.7C2 + ... + 2.7C6 + 7C7 = x = 28 - 2

Hence, the correct answer is option 2)

Concept:

The number of ways of selection from an identical object is 1.

Calculation:

It doesn’t matter which two balls you select, you’ll end up with the same selection each time. Because the balls are identical.

Therefore the number of ways to select two balls will be one.

Given:

5, 6, 7, 8, 9 are the digits to form 3 digit number

Calculation:

Let us take the 3digit number as H T U (Hundreds, tens, unit digit)  respectively

To make 3 digit number as odd

5, 7, 9 are only possibly be used in the unit digit place

In hundreds and tens place all  5 digits are possible 

Number of ways for unit digit = 3

Number of ways for tens digit = 5

Number of ways for hundreds digit = 5

Number of 3 digits odd number =  3 × 5 × 5 = 75 

∴ 75 Three-digit odd numbers can be formed from the digits 5, 6, 7, 8, 9 if the digits can be repeated

Concept:

The number of ways of selection from an identical object (Same type) is 1.

Calculation:

It doesn’t matter which four books you select, you’ll end up with the same selection each time. Because the books are same.

Therefore the number of ways to select four books will be one.

Concept:

• ⁿC_r = ⁿC_n-r.
• If nCx = nCy, then x = y or x + y = n.
• nCr + nCr-1 = n+1Cr.

Calculations:

\(\rm \begin{pmatrix} 15\\ 8\end{pmatrix}+\begin{pmatrix} 15\\ 7\end{pmatrix}=\begin{pmatrix} \rm n\\ \rm r\end{pmatrix}\)

⇒ 15C8 + 15C7 = nCr

Using nCr + nCr-1 = n+1Cr:

⇒ ​15C8 + 15C8-1 = ¹⁵⁺¹C₈ = nCr

⇒ ​16C8 = nCr

⇒ n = 16 and r = 8.

Concept:

Permutation: Permutation is defined as arrangement of r things that can be done out of total n things. This is denoted by ⁿP_r: \(\left( {{{\rm{\;}}^{\rm{n}}}{{\rm{P}}_{\rm{r}}}{\rm{\;}} = {\rm{\;}}\frac{{{\rm{n}}!}}{{\left( {{\rm{n\;}} - {\rm{\;r}}} \right)!}}} \right)\)

Arrangements of n different objects around a circle, then the number of arrangements is (n – 1)!

Calculation:

First, arrange 5 women at the round table in (5-1)! ways = 4! ways

Now, 5 space generated between the women.

Where w1 denotes the position of women and X denotes the space.

Then 3 men sit on the 5 space generated between the women which can be done in ⁵P₃ ways

Hence total number of ways is = 4! × ⁵P₃ = 24 × 60 = 1440 ways

Concept:

\(\rm ^nC_r = \dfrac{n!}{r!(n-r)!}\)

Calculations:

Given, Naresh has 10 friends, and he wants to invite 6 of them to a party.

Remove the 3 particular friends and invite 6 friends from the remaining = 10 - 3 = 7 friends.

This can be done in ⁷C₆ ways
 

Hence, required number of ways = 7C6 

⇒required number of ways = \(\rm \dfrac{7!}{(7-6)!6!}\)

⇒required number of ways = 7 

Hence, Naresh has 10 friends, and he wants to invite 6 of them to a party. total number of ways that 3 particular friends never attend the party = 7

Concept:

Basic Principle of Counting:

If there are m ways for happening of an event A, and corresponding to each possibility there are n ways for happening of event B, then the total number of different possibilities for happening of events A and B are:

• Either event A OR event B alone = m + n.
• Both event A AND event B together = m × n.

Calculation:

The 'Thousands' place can be filled by any of the 9 numbers from {1 to 9}. → 9 ways.

Since, the digits have to be distinct, the 'Hundreds' place can be filled in 9 ways (including 0). → 9 ways.

For the 'Tens' place. → 8 ways.

For the 'Units' place. → 7 ways.

Total number of possible ways of writing the four digits = 9 × 9 × 8 × 7 = 4536.

Calculation:

Odd digits - 1, 3, 5, 7, 9

A four-digit number with all digits odd has 5 choices in the first position, and since digits can be repeated, there are 5 choices for the second, third, and fourth positions also.

So, the number of four-digit natural numbers such that all of the digits are odd = 5 × 5 × 5 × 5 = 625

Concept:

• The ways of arranging n different things = n!
• The ways of arranging n things, having r same things and rest all are different = \(\rm n!\over r!\)
• The no. of ways of arranging the n arranged thing and m arranged things together = n! × m!
• The number of ways for selecting r from a group of n (n > r) = nCr 

Calculation:

There are 3 vowels O, I, E

Required number of ways N = 1 vowel and 3 consonants + 2 vowel and 2 consonants + 3 vowel and 1 consonants

⇒ N = [\(\rm\left(^3C_1×{^3C_3}\right) + \left({^3C_2}×{^3C_2}\right) + \left({^3C_3}×{^3C_1}\right)\)] × 4!

⇒ N =  (3 + 9 + 3) × 4!

⇒ N = 15 × 4! words

Concept:

Permutation: Permutation is defined as an arrangement of r things that can be done out of total n things. This is denoted by \(\rm {\;^n}{P_r}\)

\(\rm {\;^n}{P_r}\; = \;\frac{{n!}}{{\left( {n\; - \;r} \right)!}}\)

Combination: The number of selections of r objects from the given n objects is denoted by \(\rm {\;^n}{C_r}\;\) 

 \(\rm {\;^n}{C_r} = \frac{{n!}}{{r!\left( {n\; - \;r} \right)!}}\)

Note: Use combinations if a problem calls for the number of ways of selecting objects.

Fundamental Principle of Counting:

Fundamental Principle of Multiplication:

Let us suppose there are two tasks A and B such that task A can be done in m different ways following which the second task B can be done in n different ways. Then the number of ways to complete the task A and B in succession respectively is given by: m × n ways.

Fundamental Principle of Addition:

Let us suppose there are two tasks A and B such that task A can be done in m different ways and task B can be completed in n ways. Then the number of ways to complete either of the two tasks is given by: (m + n) ways.

Some of the most commonly used keywords in a word problem are shown in the table below:

Calculation:

Given: 6 coats 5 belts and 4 caps

First man can wear any of the 6 coats.

Second man can wear any of the remaining 5 coats.

Third man can wear any of the remaining 4 coats.

Therefore the number of ways in which 3 men can wear 6 coats = 6 × 5 × 4 = 120

Similarly,

Number of ways in which 3 men can wear 5 belts = 5 × 4 × 3 = 60

Similarly,

Number of ways in which 3 men can wear 4 caps = 4 × 3 × 2 = 24

Therefore, required number of ways = 120 × 60 × 24 = 172800