Permutations and Combinations 1

Hello learners, 

We are back after 6 years! Advance New Year wishes to all! 

Let's start with some Quantitative Aptitude topics! Today, let's begin with Permutations and Combinations - Day 1. 

Let's begin with the formulae! 

1. Permutation (Arrangement):

$$n{P}_r=\frac{n!}{(n-r)!}$$

Where $n$ is the total number of items, $r$ is the number of items to arrange, and $n! = n \times (n-1) \times \dots \times 1$.

2. Permutation (Arrangement):

$$n{P}_r=\frac{n!}{(n-r)!}$$

Where $n$ is the total number of items, $r$ is the number of items to arrange, and $n! = n \times (n-1) \times \dots \times 1$.

3. Factorial:

• $0! = 1$
• $n! = n \times (n-1) \times (n-2) \times \dots \times 1$

4. Repetition Permutation (when some items are identical):

$$\text{Total arrangements} = \frac{n!}{p_1! \times p_2! \times \dots \times p_k!}$$

Where $p_1, p_2, \dots, p_k$ are the frequencies of identical items.

5. Circular Permutation:

• For $n$ distinct items in a circle: $(n-1)!$
• If rotation is considered the same, divide further by $n$.

Question 1: Permutations

How many different ways can 4 people sit in a row?

Hint : Use nPr for arrangements

This is an arrangement problem, so use $n{P}_r=\frac{n!}{(n-r)!}$

Here, $n = 4$ and $r = 4$.

$$4P_4=\frac{4!}{(4-4)!}=\frac{4!}{0!}$$

Simplify the factorials.

$$4! = 4 \times 3 \times 2 \times 1 = 24,\quad 0! = 1$$
$$4P_4=\frac{24}{1}=24$$

Answer: There are 24 ways.

Question 2: Combinations

In how many ways can 3 students be selected from a group of 5 students?

Hint : Use nCr for combinations

This is a selection problem, so use $n{C}_r=\frac{n!}{r!\times (n-r)!}$

Here, $n = 5$and $r = 3$.

$$5C_3=\frac{5!}{3!\times (5-3)!}=\frac{5!}{3!\times 2!}$$

Simplify the factorials.

$$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120,\quad 3! = 3 \times 2 \times 1 = 6,\quad 2! = 2 \times 1 = 2$$
$$5C_3=\frac{120}{6\times 2}=\frac{120}{12}=10$$

Answer: There are 10 ways.

Question 3: Permutations with Repetition

How many unique arrangements can be made with the letters of the word SUCCESS?

Hint : Apply the repetition formula

This is a permutation problem with repetition, so use:

$$\text{Total arrangements}=\frac{n!}{p_1!\times p_2!\times \cdots \times p_k!}$$

Count the letters.

The word SUCCESS has 7 letters in total, with some letters repeating:

• S occurs 3 times.
• C occurs 2 times.
• U occurs 1 time.
• E occurs 1 time.

$$\text{Total arrangements}=\frac{7!}{\mathrm{3!}\times 2! \; \times \; 1! \; \times \; 1! }$$

Simplify the factorials.

7!=7×6×5×4×3×2×1=5040    3!=3×2×1=6    2!=2×1=2    1!=1

Answer: There are 420 unique arrangements of the letters in the word SUCCESS.

Question 4: Mixed (Selection + Arrangement)

A committee of 3 people is to be formed from a group of 6 people. In how many ways can the committee be formed if the positions are President, Secretary, and Treasurer?

Hint : Combine nCr​ for selection and r! for arrangement

This involves selecting and arranging, so use $n{P}_r=\frac{n!}{(n-r)!}$

Here, $n = 6$ and $r = 3$.

$$6P_3=\frac{6!}{(6-3)!}=\frac{6!}{3!}$$

Simplify the factorials.

$$6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720,\quad 3! = 3 \times 2 \times 1 = 6$$
$$6P_3=\frac{720}{6}=120$$

Answer: There are 120 ways.

Question 5: Circular Permutation

In how many ways can 7 friends sit around a circular table?

Hint : Apply the circular permutation formula

For circular arrangements of $n$ items, use:

$$(n-1)!$$

Here, $n=\mathrm{7:}$

$$(7-1)!=6!$$

Simplify the factorial.

$$\mathrm{<span\; class="base"><span\; class="mord">6</span><span\; class="mclose">!</span><span\; class="mspace"></span><span\; class="mrel">=</span><span\; class="mord">6</span><span\; class="mspace"></span><span\; class="mbin">\times </span><span\; class="mord">5</span><span\; class="mspace"></span><span\; class="mbin">\times </span><span\; class="mord">4</span><span\; class="mspace"></span><span\; class="mbin">\times </span><span\; class="mord">3</span><span\; class="mspace"></span><span\; class="mbin">\times </span><span\; class="mord">2</span><span\; class="mspace"></span><span\; class="mbin">\times </span><span\; class="mord">1</span><span\; class="mspace"></span><span\; class="mrel">=</span>720</span>}$$

Answer: There are 720 ways.

Have fun with permutations and combinations—arranging and selecting in endless ways!

Quants 26 - Interests

 Practice Questions:

Question 1:

What is the difference between the simple interest on a principal of ` 500 being calculated at 5% per annum for 3 years and 4% per annum for 4 years?

5% for 3 years (SI) = 15% of the amount; At the same time 4% SI for 4 years means 16% of the amount. The difference between the two is 1% of the amount. 1% of 500 = ` 5

(or)

SI = PNR/100 = (500*5*3)/100 = 75

SI = PNR/100 = (500*4*4)/100 = 80

Difference = 80 - 75 = 5.

Question 2:

In what time will ` 3300 become ` 3399 at 6% per annum interest compounded half-yearly?

Since compounding is half yearly, it is clear that the rate of interest charged for 6 months would be 3% 

3300 -3%- 3399.

Question 3:

What is the simple interest for 9 years on a sum of ` 800 if the rate of interest for the first 4 years is 8% per annum and for the last 4 years is 6% per annum?

8% of 800 for 4 years + 6% of 800 for 4 years = 64 × 4 + 48 × 4 = 256 + 192 = 448. 

However, we do not know the rate of interest applicable in the 5th year and hence cannot determine the exact simple interest for 9 years.

Question 4:

What is the difference between compound interest and simple interest for the sum of ` 20,000 over a 2 year period if the compound interest is calculated at 20% and simple interest is calculated at 23%?

Simple interest @ 23% = 4600 × 2 = 9200
Compound interest @ 20%

20000 -20%- 24000 -20%- 28800 Æ ` 8800 compound interest.
Difference = 9200 – 8800 = ` 400.

Question 5:

Find the compound interest on ` 1000 at the rate of 20% per annum for 18 months when interest is compounded half-yearly.

1000 -10%- 1100 -10%- 1210 -10%- 1331.
Compound interest = 1331 – 1000 = ` 331

Time to Think:

What is the simple interest on a sum of `700 if the rate of interest for the first 3 years is 8% per annum and for the last 2 years is 7.5% per annum?

Quants 25 - Interest

Practice Questions:

Question 1:

` 1200 is lent out at 5% per annum simple interest for 3 years. Find the amount after 3 years.

The annual interest would be ` 60 (PNR/100 = (1200*1*5)/100). 

After 3 years the total value would be 1200 + 60 × 3 = 1380.

Question 2:

Interest obtained on a sum of ` 5000 for 3 years is ` 1500. Find the rate percent.

The interest earned per year would be 1500/3=500. This represents a 10% rate of interest.

SI = PNR/100 => 1500 = (5000 x 3 x R)/100 => R = 1500 x 100 / ( 5000 x 3 ) = 10%.

Question 3:

` 2100 is lent at compound interest of 5% per annum for 2 years. Find the amount after two years.

2100 + 5% of 2100 = 2100 + 105 = 2205 (after 1 year). 

Next year it would become:
2205 + 5% of 2205 = 2205 +110.25 = 2315.25

Question 4:

Find the difference between the simple and the compound interest at 5% per annum for 2 years on a principal of ` 2000.

Simple Interest for 2 years = 100 + 100 = 200.
Compound interest for 2 years: 

Year 1 = 5% of 2000 = 100.
Year 2: 5% of 2100 = 105 Æ Total compound interest = ` 205.
Difference between the Simple and Compound interest = 205 – 200 = ` 5

Question 5:

After how many years will a sum of ` 12,500 become ` 17,500 at the rate of 10% per annum?

12500 @ 10% simple interest would give an interest of ` 1250 per annum. For a total interest of ` 5000, it would take 4 years.

Time to Think:

Find the rate of interest if the amount after 2 years of simple interest on a capital of ` 1200 is ` 1440.

This week and the following week is a special week for Quantitative aptitude. Full fledged to concepts in 10 days!! Happy learning!!

Answers Week 12

Quants 23 - Stocks and Shares

A 9% stock yields 8%. The market value of the stock is:

To obtain Rs.8, investment, Rs. 100

To obtain Rs.9, investment, Rs. (100/8) * 9 = Rs.112.50

Market value of Rs. 100 stock = Rs.112.50

 

Logical 23 - Classification

a) Pistol

b) Sword

c) Gun

d) Rifle

e) Cannon

Here, all except Sword are fire arms, and can be used from a distance.

a) Cathedral

b) Mosque

c) Church

d) Monastery

e) Temple

All except Monastery are places of worship, while monastery is a place where monks stay.


Verbal 12 - Sentence Completion

We have to bear our own burdens.

Napoleon was killed in the battle of Waterloo.

The world is on the verge of a third world war.

Barter is an economic transaction involving exchange of articles.

The premises is for sale.

Logical 24 - Verification of Truth of the Statement

A factory always has

a) Electricity

b) Chimney

c) Workers

d) Files

e) Sellers

A clock always has

a) Battery

b) Numbers

c) Alarm

d) Needles

e) Frame


Quants 23 - Stocks and Shares

A can do a work in 15 days and B in 20 days. If they work together for 4 days, then the fraction of the work left is:

A's 1 day work = 1/15

B's 1 day work = 1/20
A+B's 1 day work = 1/15 + 1/20 = 7/60
A+B's 4 days work = 7/60 * 4 = 7/15
Thus, remaining work = 1 - 7/15 = 8/15

Quants 24 - Time and Work

Practice Questions:

Question 1:

Sakshi can do a piece of work in 20 days. Ziva is 25% more efficient than Sakshi. The number of days taken by Ziva to do the same piece of work is:

Ratio of times taken by Sakshi and Ziva is 125:100 = 5:4
Suppose Ziva takes x days to do the work,
5 : 4 :: 20 : x => x = (4 * 20) / 5 = 80/5 = 16 days.
Hence, Ziva takes 16 days to complete the work.

Question 2:

A is 30% more efficient than B. How much time will they, working together, take to complete a job which A alone could do in 23 days?

Ratio of times taken by A and B = 100:130 = 10:13
Suppose B takes x days to do the work,
Then 10 : 13 :: 23 : x => x = (23 * 13) / 10 = 299/10
A's 1 day work = 1/23; B's 1 day work = 10/299
A+B 's 1 day work = 1/23 + 10/299 = 23/299 = 1/13
A and B together can complete the work in 13 days.

Question3:

A does half as much as B in three-fourth of the time. If together, they take 18 days to complete the work, how much time shall B take to do it?

Suppose B takes x days to complete the work,
A takes 2*(3/4)x = 3/2 x days
A+B 's 1 day work = 1/18
1/x + 2/3x = 1/18
x = 36/5 days.

Question 4:

A is 50% as efficient as B. C does half of the work done by A and B together. If C alone does the work in 40 days, then A, B and C together can do the work in:

A's 1 day work : B's 1 days work = 150 : 100 = 3 : 2
Let A's and B's 1 day work be 3x and 2x respectively.
Then C's 1 day work = (3x+2x)/2 = 5x/2
5x/2 = 1/40 => 5x = 1/20 => 100x = 1
x= 1/100
A's 1 day work = 3/100
B's 1 day work = 2/100 = 1/50
C's 1 day work = 5/100 / 2 = 5/200 = 1/40
A+B+C 's 1 day work = 3/100 + 2/100 + 5/200 = 15/200 = 3/40
So, A, B and C can complete the work in 40/3 = 13 1/3 days.

Question 5:

Two workers A and B working together completed a job in 5 days. If A worked twice as efficiently as he actually did and B worked 1/3 as efficiently as he actually did, the work would have been completed in 3 days. A alone could complete the work in:

Let A's 1 day work = x
Let B's 1 day work = y
Then x+y = 1/5 and 2x + 1/3y = 1/3
Solving the above 2 equations,
x = 4/25 and y = 1/25
A's 1 day work = 4/25
So, A alone could complete the work in 25/4 = 6 1/4 days.

Time to Think:

A can do a work in 15 days and B in 20 days. If they work together for 4 days, then the fraction of the work left is:

Logical 24 - Verification of Truth of the Statement

Key Points to Remember:

In this type of questions, we are required to stress only on the truth of the facts that always hold. Questions are asked in context of a particular thing or factor that is always characterized by a specific part or feature. The alternatives other than the correct answer also seem to bear a strong relationship with the thing mentioned. So, the absolute truth is to be followed. 

Examples:

1. Atmosphere always has

a) Oxygen

b) Air

c) Germs

d) Moisture

e) Dust

Clearly, though all the alternatives may form a part of the atmosphere, the air is the most vital part, without which there can be no atmosphere. 

2. A train always has

a) Engine

b) Rails

c) Driver

d) Guard

e) Passengers

Clearly, rails are necessary for the train to move on. Driver alone can move the train. A guard is also necessary for safety. A train is moved for the passengers. But all these do not constitute a train. A train cannot be called so without the engine. 

3. Which one of the following is always found in Bravery?

a) Courage

b) Experience

c) Power

d) Knowledge

Clearly, bravery is a quality exhibited only by a person who possesses courage. 

Practice Questions:

4. Which of the following an animal always has?

a) Lungs

b) Skin

c) Mind

d) Heart

e) Life

5. A race always has

a) Referee

b) Spectators

c) Rivals

d) Prize

e) Victory

6. Which of the following a Drama must have?

a) Actors

b) Story

c) Sets

d) Director

e) Spectators

7. A book always has

a) Chapters

b) Pages

c) Contents

d) Pictures

e) Illustrations

8. A mirror always

a) Reflects

b) Retracts

c) Distorts

d) Refracts

e) Reveals the truth

Time to Think

9. A factory always has

a) Electricity

b) Chimney

c) Workers

d) Files

e) Sellers

10. A clock always has

a) Battery

b) Numbers

c) Alarm

d) Needles

e) Frame