## 1st Bank PO level Solution Set, 1st on topic Permutation and Combination

This is the 1st solution set of 10 practice problem exercise for Bank PO exams and 1st on topic Permutation and Combination. Students must complete the corresponding question set in prescribed time first and then only refer to this solution set for gaining maximum benefits from this resource.

In MCQ test, you need to deduce the answer in shortest possible time and select the right choice.

Based on our analysis and experience we have seen that, for accurate and quick answering, the student

- must have complete understanding of the basic concepts in the topic area
- is adequately fast in mental math calculation
- should try to solve each problem using the basic and rich concepts in the specific topic area and
- does most of the deductive reasoning and calculation in his or her head rather than on paper.

Actual problem solving is done in the fourth layer. You need to use **your problem solving abilities** to gain an edge in competition.

We list below the few important formulas for permutation and combination. If you already know, you may skip.

### Basic formulas on permutation and combination

**First case:** **Permutation** (or number of possible ordered arrangements) of $r$ **distinct objects** out of $N$ distinct objects. It is,

$^NP_r=\displaystyle\frac{N\text{!}}{(N-r)\text{!}}$, where Factorial $N$, expressed as $N\text{!}$, is equal to the product of integers starting from $N$ decreasing by 1 and ending with 1.

In other words,

$N!=N\times{(N-1)}\times{(N-2)}\times{...}\times{2}\times{1}$.

With this knowledge we can express,

$^NP_r=N\times{(N-1)}\times{(N-2)}\times{...}\times{(N-r+1)}$, cancelling out $(N-r)\text{!}$ between numerator and denominator.

For example, permutation of 3 distinct objects out of 8 distinct objects will be,

$^8P_3=8\times{7}\times{6}$, the denominator $(8-3)\text{!}=5\text{!}$ is cancelled out with part of the numerator.

**Second case: ** **Permutation** of $r$ objects out of $N$ objects with $q$ objects in the set of $N$ **objects alike** or same,

$^NP_r\text{ with q alike}=\displaystyle\frac{N\text{!}}{(N-r)\text{!}\times{q\text{!}}}$.

**Third case:** **Combination or selection** of $r$ distinct objects out of $N$ distinct objects,

$^NC_r=\displaystyle\frac{N\text{!}}{r\text{!}\times{(N-r)\text{!}}}={^NC_{N-r}}$.

For more details and clear understanding of the concepts, you should refer to our extensive tutorial,

* Permutation and Combination with exercises*.

If you have not taken the paired test yet, you should take the test at * Bank PO level Quantitative aptitude question set 1 on permutation and combination 1*, and then only go through the following solutions for appreciating the solutions fully.

### 1st solution set - 10 problems for Bank PO exams: 1st on topic Permutation and Combination - time 12 mins

**Problem 1.**

Out of 5 women and 4 men, a committee of three members is to be formed in such a way that at least one member is a woman. In how many different ways can it be done?

- 76
- 80
- 84
- 96
- None of the above

**Solution 1: Problem analysis and solution by problem modelling**

In permutation and combination problems, the most important task is to analyze the problem and form the right problem model. Once you do that, rest is easy. Just apply the formulas of permutation or combination in the problem model.

This is a group formation problem where the group is a committee.

The solution to the problem here can be achieved in two ways; these paths to the solution we call as problem models.

Let's start with the easiest one.

**First problem model:** at least one woman in the committee means—committee formations with **Not ALL MEN.**

We will then find out the number of total selections of 3 out of **total 9 men and women** and then **subtract all possible combinations of 3 out of 4 men** (ALL MEN committees). That will give us all selections of the 3 member committee with at least a woman as,

$^9C_3-{^4}C_3$

$=\displaystyle\frac{9\times{8}\times{7}}{3\times{2}}-4$

$=84-4$

$=80$.

We have just subtracted the "ALL MEN" possibilities from total number of possible committees so that it is ensured that at least one woman gets a place in every possible committee.

This is the **problem solving approach** resulting in **quickest solution** with *least amount of calculations*.

We can name this model as, "At least one model" or "At least one method".

**Second problem model:** Enumerate each possible number of men and women formation, and add combinations for each formation.

There are three possible formations of the committee,

- 1 woman out of 5, and 2 men out of 4,
- 2 women out of 5, and 1 man out of 4, and
- 3 women out of 5.

These three formations will cover all combinations of the committee with at least 1 woman.

Number of combinations of the three formations are respectively,

$^5C_1\times{^4C_2}=5\times{6}=30$,

$^5C_2\times{^4C_1}=10\times{4}=40$, and

$^5C_3=10$.

The three sums up to 80.

This is the more time-consuming approach.

**Answer.** Option b: 80.

**Key concepts used:** * Problem modelling* --

*--*

**Permutation and combination***--*

**Problem solving approach***.*

**Shortcut solution -- At least one method -- Enumeration technique -- Group formation problem -- Solving in mind -- Negation technique****Note:** The first solution could be reached wholly in mind in a few tens of seconds.

**Problem 2.**

In how many different way can the letters of the word SOFTWARE be arranged in such a way that the vowels always come together?

- 13440
- 120
- 1440
- 360
- None of the above

**Solution 2: Problem analysis and solution by problem modelling**

**First conclusion:** This is a permutation problem (as it is arrangement of letters and orders in which the letters appear in an arrangement is important). The number of letters is 8. All letters are distinct with no repetition, but in desired permutations, all three vowels O, A and E must appear together in a cluster.

Let us go ahead with describing and explaining the right problem model.

**Appearing together model or method:** The technique of dealing with this type of problem is to go through,

**Compaction phase:**Consider 3 vowels as 1**merged vowel**, represented by the symbol say, "+". The reduced number of objects will then be 6 and all 6 will be unique.- Evaluate then all possible permutations of 6 out of 6. This is all possible arrangements with 3 vowels as 1 merged vowel.
**Expansion:**Evaluate all possible permutations of merged vowels 3 out of 3, which will be 6. We are now expanding each merged vowel to 6 possible arrangements between themselves without separating them from each other.**For each 6 out of 6 permutations there will then be 6 permutations of the merged vowels when expanded in the position they appear.****Final evaluation:**Evaluate the total desired number of permutations as a product of 6 out of 6 permutations and 3 out of 3 permutations.

It will then be,

$^6P_6\times{^3P_3}$

$=6\text{!}\times{3\text{!}}$, by definition $0\text{!}=1$,

$=6\times{5}\times{4}\times{3}\times{2}\times{6}$

$=36\times{120}$

$=4320$.

**Answer:** Option e: None of the above.

**Key concepts used:** * Permutation and combination* --

*Problem modelling -- Appearing together model.*For accuracy we wrote down the factors of the final product, though evaluation was easy and quick.

#### Explanation of appearing together problem model with an example

Say a unique arrangement among the 6 out of 6 arrangements is, "FTS+RW", where "+" represents the merged state of three vowels A, E and O. In this particular arrangement, the merged vowel appears in 4th position.

When the **merged vowel** is expanded in 3 out of 3 or 6 ways, the **single unique original arrangement of six objects** will **generate 6 unique arrangements of 8 letters**, with the three vowels appearing in 4th, 5th and 6th positions, and still together. These six arrangements will be,

FTSAEORW, FTSAOERW, FTSOAERW, FTSOEARW, FTSEAORW and FTSEOARW.

This way all of 6 out of 6 unique arrangements will generate 6 times 6 out of 6, or, 4320 numbers of unique 8 letter arrangements where three vowels will always appear together.

**Problem 3.**

A team of 5 children is to be selected out of 4 girls and 5 boys so that it contains at least 2 girls. In how many ways the selection can be made?

- 105
- 120
- 60
- 100
- None of the above

**Solution 3: Problem analysis and solution by problem modelling**

**First model:** The desired selections will contain only 2, or 3 or 4 girls. This is enumeration technique—easier to visualize but time-consuming to evaluate.

By this model the desired number will be,

$^4C_2\times{^5C_3}+{^4}C_3\times{^5C_2}+{^4}C_4\times{^5C_1}$

$=6\times{10}+4\times{10}+1\times{5}$

$=60+40+5$

$=105$.

You need to be careful in evaluations.

**Second model:** **At least two method**—a variation of *at least one method*: From all selections of 5 members out of total 9 girls and boys, we will subtract selections for ALL BOYS and selections with exactly 1 girl. This might be a bit more difficult to visualize, but easier to calculate.

The desired number of selections by this model is then,

$^9C_5-{^5}C_5-{^4}C_1\times{^5C_4}$

$=126-1-4\times{5}$

$=105$.

Choose your method. We use the second method that is more conceptual with quick evaluation. It can be done easily in mind. This represents the problem solving approach.

**Answer:** Option a: 105.

**Key concepts used:** * Problem modelling* --

*.*

**Permutation and combination --****Problem solving approach -- Group formation problem -- Enumeration technique -- At least one method -- Adapted At least one method -- Shortcut solution -- Negation technique****Problem 4.**

In how many different ways a group of 4 men and 4 women be formed out of 7 men and 8 women?

- 105
- 2450
- 1170
- Cannot be determined
- None of the above

**Solution 4: Problem analysis and solution**

This is a group formation problem of selecting 4 out of 7 men, 4 out of 8 women and taking the product of the two.

The desired number of different ways such a group can be formed is,

$^7C_4\times{^8C_4}$

$=\displaystyle\frac{7\times{6}\times{5}\times{4}}{4\times{3}\times{2}}\times{\displaystyle\frac{8\times{7}\times{6}\times{5}}{4\times{3}\times{2}}}$

$=35\times{70}$

$=2450$.

For each of the $^7C_4$ selections of men, there will be $^8C_4$ number of selections for women, and vice versa. That's why the total number ways the group can be formed is given by the product of these two.

**Answer:** Option b: 2450.

**Key concepts used:** **Group formation problem -- Permutation and combination***.*

**Problem 5.**

On a shelf there are 3 books on Management, 4 books on Economics and 4 books on Statistics. In how many different ways the books can be arranged so that the books on Economics are kept together?

- 5040
- 120960
- 40320
- 967680
- None of the above

#### Solution 5: Problem analysis and solution by using appearing together method

We will use the appearing together method in this permutation problem.

Merging 4 books on Economics into 1, number of ways distinct 8 number of objects can be permuted out of 8 is,

$^8P_8$.

The number of distinct arrangements of 4 Economics books within themselves as a group is,

$^4P_4$.

So total number of different ways the books can be arranged so that Economics books are always kept together is,

$^8P_8\times{^4P_4}$

$8\text{!}\times{4\text{!}}$

$=56\times{30}\times{24}\times{24}$, we write down products of two consecutive numbers,

$=168\times{576}$

$=967680$.

**Answer:** Option d: 967680.

**Key concepts used:** * Permutation and combination* --

**Appearing together method.**We had to calculate the last product by hand.

#### Problem 6.

A committee of 5 members is to be formed out of 4 students, 3 teachers and 2 sports coaches. In how many ways can the committee be formed if the committee should consist of 2 students, 2 teachers and 1 sports coach?

- 25
- 9
- 64
- 36
- None of the above

**Solution 6: Problem analysis and solution**

It is a straightforward selection of three subsets, 2 students out of 4, 2 teachers out of 3, 1 sports coach out of 2, and taking the product of the three,

$^4C_2\times{^3C_2}\times{^2C_1}$

$=6\times{3}\times{2}=36$.

**Answer:** Option d: 36.

**Key concepts used:** **Permutation and combination -- Problem modelling -- Group formation problem**** -- Solving in mind.**

The problem could easily be solved in mind.

**Problem 7.**

Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels be formed?

- 24400
- 210
- 21300
- 25200
- None of the above

**Solution 7: Problem analysis and Problem solving in two phases**

This is a problem consisting of two parts.

**First part:** we need to find the product of two selections, 3 consonants out of 7 and 2 vowels out of 4. These will be the unique combinations of 3 consonants and 2 vowels,

$^7C_3\times{^4C_2}$

$=\displaystyle\frac{7\times{6}\times{5}}{3\times{2}}\times{\displaystyle\frac{4\times{3}}{2}}$

$=35\times{6}$

$=210$.

This is the Combination part.

**Second part:** Now we need to permute each of these unique 5 letter combinations in $^5P_5=5\text{!}$ ways to generate all possible words consisting of selections of 3 consonants out of 7 and 2 vowels out of 4. This is the permutation part and the desired number of words will then be,

$210\times{5}\times{4}\times{3}\times{2}$

$=210\times{120}$.

**Answer:** Option d: 25200.

**Key concepts used:** * Permutation and combination mixed* --

*.*

**Two part group formation problem****Note:** Forming words will always involve permutation.

**Problem 8.**

A committee of 12 persons is to be formed from 9 women and 8 men. In how many ways this can be done if at least 5 women have to be included in the committee?

- 6000
- 6005
- 6010
- 6062
- None of the above

**Solution 8: Problem analysis, Problem modelling and solution by adapted at least one method**

By adapting "**at least one**" concept to this problem, we should get the desired number of selections by subtracting selections with exactly 0, 1, 2, 3, and 4 women from all combinations to get the selections with at least 5 women. But as there are only 8 men and committee strength is to be 12, we will just have to subtract selections with exactly 4 women, rest four being 0.

So desired number of selections with at least 5 women will be,

$^{17}C_{12}-{^8C_8}\times{^9C_4}$

$={^{17}C_5}-1\times{\displaystyle\frac{9\times{8}\times{7}\times{6}}{24}}$, as $^NC_r={^NC_{N-r}}$,

$=\displaystyle\frac{17\times{16}\times{15}\times{14}\times{13}}{5\times{4}\times{3}\times{2}}-9\times{14}$

$=17\times{2}\times{14}\times{13}-9\times{14}$

$=(34\times{13}-9)\times{14}$

$=433\times{14}$

$=6062$.

**Answer:** Option d: 6062.

**Key concepts used:** **Problem analysis -- Problem modelling -- Key pattern identification -- Adapted at least one method -- Permutation and combination -- Efficient simplification -- Negation technique.**

Identifying the key pattern that only one selection with exactly 4 women need to be subtracted simplified the evaluation considerably.

**Problem 9.**

Two girls and 4 boys are to be seated in a row in such a way that the girls do not sit together. In how many different ways can it be done?

- 360
- 720
- 480
- 240
- None of the above

**Solution 9: Problem analysis, Problem modelling and solution by Adapted Appearing together method**

This is a permutation problem.

By **Appearing together method** we can easily find out the number of ways in which two girls always sit together. It will be,

$^5P_5\times{2}=20\times{6}\times{2}=240$.

If we subtract this number from the total number of sitting arrangements with no restriction, we will get the number of sitting arrangements in which no two girls sit together,

$^6P_6-240=30\times{24}-240=480$

**Answer:** Option c: 480.

**Key concepts used:** * Problem analysis* --

*--*

**Problem modelling***--*

**Adapted appearing together method**

**Permutation and combination --**

**Permutation -- Negation technique.****Problem 10.**

In how many different ways can the letters in the word BANKING be arranged?

- 2520
- 5040
- 5080
- 2540
- None of the above

**Solution 10: Problem analysis and solution by pattern identification and problem modelling**

This is 7 out of 7 permutation with a twist. The key pattern identified is the occurrence of two "N"s in the word BANKING.

So to get the unique arrangements of the letters in the word, we need to divide $^7P_7=7{!}$ by $2\text{!}$, giving,

$\displaystyle\frac{7\times{6}\times{5}\times{4}\times{3}\times{2}}{2}$

$=7\times{360}$

$=2520$.

**Answer.** Option a: 2520.

**Key concepts used:** * Key pattern identification* --

*--*

**Problem modelling***.*

**Permutation with alike objects**### Further reading resources

**Tutorial on Permutation and Combination with exercise problems**

**Solutions 1 to the exercise problems on Permutation and combination**

**Bank PO level Solution set 2 on Permutation and Combination 2**

**Bank PO level Question set 2 on Permutation and Combination 2**

**Bank PO level Quantitative aptitude Solutions 1 on Permutation combination 1**

**Bank PO level Quantitative aptitude Question set 1 on Permutation combination 1**