Bank PO level Quantitative aptitude Solutions 2 on Permutation and Combination 2

Submitted by Atanu Chaudhuri on Tue, 20/11/2018 - 15:49

2nd Bank PO level Solution Set, 2nd on topic Permutation and Combination

Bank PO Solutions 2 permutation combination 2

This is the 2nd solution set of 10 practice problem exercise for Bank PO exams and 2nd 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,

${^N}P_r=\displaystyle\frac{N\text{!}}{(N-r)\text{!}}$, 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,

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

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

${^8}P_3=8\times{7}\times{6}$, 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,

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

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

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

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 2 on permutation and combination 2, and then only go through the following solutions for appreciating the solutions fully.

2nd solution set - 10 problems for Bank PO exams: 2nd on topic Permutation and Combination - time 12 mins

Problem 1.

In how many different ways can the digits in the number "256974" be arranged using each digit only once in each arrangement such that the digits 6 and 5 are at the extreme end of the arrangements?

36
360
48
720
None of the above

Solution 1: Problem analysis and solution by problem modelling

Out of 6 digits in the number "256974" when we fix the positions of 6 and 5 at two extreme ends,

the number of digits left is reduced to 4 as well the positions in which these can be arranged is reduced to 4. These are the 4 positions between two extreme ends.

The different ways these 4 digits can be arranged is,

${^4}P_4=4\text{!}=24$.

For each of these 24 arrangenents the digits 6 and 5 will be at two extreme ends, but in two ways, 6-5 and 5-6.

So total number of different arrangements as asked for is,

$2\times{24}=48$.

Answer. Option c: 48.

Key concepts used: Problem modelling -- Permutation and combination -- Permutation in parts -- Solving in mind.

Note: The solution could be reached wholly in mind in a few tens of seconds.

Problem 2.

In how many different ways can the letters of the word "CORPORATION" be arranged such that the vowels always come together?

840
8400
1440
86400
None of the above

Solution 2: Problem analysis and problem modelling

First conclusion: This is a permutation problem (as it is arrangement of letters and order in which the letters appear in an arrangement is important). The number of letters is 11. There are two barriers to evaluate a permutation of 11 out of 11,

All letters are not distinct, there are two sets of repetitions in 2 numbers of R and three numbers of O.
All 5 vowels—3 numbers of O's and A and I must come together, in a bunch of 5.

This is a mixed hybrid model of two components—the appearing together model and object repetition model (or permutation with alike objects).

Solution 2: Strategy of solving mixed type of permutation problem

As the appearing together condition is a more stringent one, it needs to be satisfied first, and then the letter repetition pattern will be applied.

Solution 2: Solving the mixed permutation problem

Appearing together model mixed with repetition: The technique of dealing with this type of problem is to go through,

Compaction phase: Consider 5 vowels as 1 merged vowel, represented by the symbol say, "+". The reduced number of objects will then be 7 and 2 R's are repeated in 7 objects. This number of permutations will be, $\displaystyle\frac{7\text{!}}{2\text{!}}=42\times{60}=2520$. This is then all possible permutations of 7 out of 7 with 2 object repetition where 5 vowels are merged into 1 merged vowel.
Expansion: Evaluate all possible permutations of merged vowels 5 out of 5 with three O's repeated. This will be, $\displaystyle\frac{5\text{!}}{3\text{!}}=20$. We are now expanding each merged vowel to 20 possible arrangements between themselves without separating them from each other. For each 7 out of 7 permutation with repetition of 2 R's there will then be 20 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 the above two permutations.

It will then be,

$2520\times{20}=50400$.

Answer: Option e: None of the above.

Key concepts used: Permutation and combination -- Problem modelling -- Appearing together model -- Object repetition model -- Permutation with alike objects -- Mixed permutation model -- Permutation in parts.

Explanation of appearing together problem model with an example

Say a unique arrangement among the 7 out of 7 arrangements is, "C+RPRTN", where "+" represents the merged state of 5 vowels. In this particular arrangement, the merged vowel appears in 2nd position.

When the merged vowel is expanded in 5 out of 5 with three vowel repetitions, or 20 number of ways, the single unique original arrangement of seven objects will generate 20 unique arrangements of 11 letters, with the 5 vowels still together.

This way all of 7 out of 7 unique arrangements with two R's repeated will generate 2520 times 20, or, 50400 numbers of unique 11 letter arrangements of the letters of the original word "CORPORATION".

Problem 3.

A committee of 12 persons is to be formed out of 9 women and 8 men. In how of these possible committees, women will be in majority?

2702
2705
2000
2700
None of the above

Solution 3: Problem analysis and solution by problem modelling

For women to be in majority in the 12 member committee, the three possible groups of combinations would have, 9 women 3 men, 8 women 4 men and 7 women 5 men. These are the only three groups of selections possible for women to be in majority. This is enumeration technique.

By this model the desired number will be,

$^9C_9\times{^8C_3}+{^9}C_8\times{^8C_4}+{^9}C_7\times{^8C_5}$

$=1\times{56}+9\times{70}+36\times{56}$, as ${^9}C_8={^9}C_1$, ${^9}C_7={^9}C_2$, and ${^8}C_5={^8}C_3$, each of which is much easier to evaluate

$=56+630+2016$

$=2702$

Answer: Option a: 2702.

Key concepts used: Problem modelling -- Permutation and combination -- Group formation problem -- Enumeration technique -- Combination in parts -- Problem breakdown technique -- Efficient simplification.

Problem 4.

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

45
120
24
126
None of the above

Solution 4: Problem analysis and solution

With no restriction on selection of committee members, the problem is transformed to all combinations of 5 out of total 9 persons,

$^9C_5=\displaystyle\frac{9\times{8}\times{7}\times{6}\times{5}}{5\times{4}\times{3}\times{2}}=126$

Answer: Option d: 126.

Key concepts used: Group formation problem -- Permutation and combination -- Any selection model -- Solving in mind.

The problem could easily solved in mind.

Problem 5.

From a group of 6 men and 4 women a committee of 4 persons is to be formed. In how many different ways can it be done so that the committee has at least one woman?

210
185
195
225
None of the above

Solution 5: Problem analysis and solution by using at least one model

For quick conceptual solution, we will use the at least one method in this combination problem.

At least one method:

If we subtract the number of combinations with ALL MEN from all possible combinations for selecting any 4 out of a total of 10 men and women, in every such combination there will always be at least one woman,

$^{10}C_4-{^6}C_4$

$=\displaystyle\frac{10\times{9}\times{8}\times{7}}{24}-{^6}C_2$, as ${^N}C_{N-r}={^N}C_r$

$=210-15$

$=195$.

Answer: Option c: 195.

Key concepts used: Permutation and combination -- At least one method -- Problem solving approach -- Group formation problem -- Solving in mind.

The problem could have been solved by enumeration of possible groups of selections, but that would have taken longer time. The at least one method enables us to solve the problem mentally, as this is the problem solver's approach.

Problem 6.

A committee of 6 teachers is to be formed out of 5 arts teachers, 4 science teachers and 3 commerce teachers. In how many ways can the committee be formed if no teacher from the commerce stream be included in any committee?

81
46
62
84
None of the above

Solution 6: Problem analysis and solution

It is a straightforward selection of 6 teachers out of a total of 9 arts and science teachers,

${^9}C_6$

$={^9}C_3$

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

$=84$.

Answer: Option d: 84.

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

The problem could easily be solved in mind.

Problem 7.

A committee of 5 members is to be formed out of 3 trainees, 6 research associates and 4 professors. In how many different ways can this be done if the committee should have all the 4 professors and 1 research associate or all 3 trainees and 2 professors?

15
12
19
25
None of the above

Solution 7: Problem analysis and Problem solving: Sum of two groups of possibilities

This is a problem consisting of two parts that are clearly specified, and not a result of problem breakdown into two parts by us.

First part: Committees consisting of all 4 professors and 1 research associate.

The number of such committees is,

${^4}C_4\times{^6C_1}=6$.

Second part: Committees consisting of all three trainees and 2 professors.

This number is,

${^3}C_3\times{^4C_2}=6$.

A total of 12 possible committe formations with given restrictions.

Answer: Option b: 12.

Key concepts used: Combination in parts -- Two part group formation problem -- Solving in mind.

Problem 8.

In how many different ways can the letters of the word "THERAPY" be rearranged so that the vowels never come together?

1440
720
5040
4800
3600

Solution 8: Problem analysis and Problem modelling

We will first have to find out the number of rearrangements where the vowels appear together always. If we subtract this number from the whole set of all possible rearrangements of the letters without any restriction, we will be left with all rearrangements where the vowels never appear together. This is negation technique applied on appearing together model.

Solution 8: Solution by appearing together method and negation technique

The word "THERAPY" has 7 letters out of which 2 are vowels. Merging these two vowels we will get 6 objects and number of unique 6 out of 6 arrangements as,

${^6}P_6=6\text{!}=720$.

Now when we expand the two merged vowels E and A in place, for each of 720 distinct arrangements we will get 2 possible arrangements.

So total number of arrangements in which the two vowels always appear together is,

$720\times{2}$.

Now we will negate or subtract this subset from the whole set of rearrangemets of 7 out of 7 arrangements, to get the arrangements with vowels never appearing together as,

$=7\text{!}-720\times{2}$

$=(7-2)\times{720}$, as $6\text{!}=720$

$=5\times{720}$

$=3600$.

Answer: Option e: 3600.

Key concepts used: Problem analysis -- Problem modelling -- Key pattern identification -- Appearing together method -- Permutation and combination -- Efficient simplification -- Negation technique.

Key pattern is transforming the "vowels never appearing together" to "all possible permutations, the whole set" minus "vowels always appearing together subset" using set negation or subtraction concept.

Problem 9.

From a group of 7 men and 6 women, 5 persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can this be done?

564
735
645
756
None of the above

Solution 9: Problem analysis and Problem modelling

The possible groups of combinations will consist of, 3 men 2 women, 4 men 1 woman and 5 men. These are the only possible committee compositions. Adding the number of selections for each group we will get the desired number of selection with at least 3 men in each selection of the committee.

Solution 9: Solution by combination in parts

Thus the number of desired selections will be,

${^7}C_3\times{{^6}C_2}+{^7}C_4\times{{^6}C_1}+{^7}C_5$

$=35\times{15}+35\times{6}+21$, as ${^7}C_4={^7}C_3$, and ${^7}C_5={^7}C_2$

$=36\times{21}$

$=756$.

Answer: Option d: 756.

Key concepts used: Problem analysis -- Problem modelling -- Permutation and combination -- Combination in parts -- Efficient simplification.

To speed up calculation we took first 35 as the common factor and added up 15 and 6 to get second factor as 21. Again we took 21 as common factor and added up 35 and 1 to result in 36 as the second factor. Ultimately instead of three multiplications we did only one multiplication at the last stage. This is what we call Efficient simplification.

Problem 10.

4 boys and 3 girls are to be seated in a row in such a way that no two boys sit adjacent to each other. In how many different ways can this be done?

72
5040
144
30
None of the above

Solution 10: Problem analysis and problem modelling

This is a permutation problem on seating arrangement. In this special model every arrangement must have a girl between two boys.

In total there being 7 positions then, the boys and girls must be seated alternately with two boys sitting at two ends. The three gaps between 4 boys will be taken up by 3 girls. This is the only possible way the boys and the girls can be seated with no two boys sitting adjacent to each other.

The 4 boys will then form a group occupying fixed positions as well as the 3 girls will form a group occupying three fixed positions.

Permutations for each group between themselves will be for boys,

${^4}P_4=4\text{!}=24$, and

For girls,

${^3}P_3=3\text{!}=6$.

Total number of distinct ways of such seating will be product of the two,

$24\times{6}=144$.

Answer. Option c: 144.

Key concepts used: Key pattern identification -- Problem modelling -- Alternate positioning model -- Permutation in groups -- Visualization.

Note: Relative positions being fixed, permutations are done within each group and then the product of the two produced the final result. This is alternate positioning model. One needs to visualize this key pattern for satisfying the given criterion of no two boys sitting adjacent to each other.

Recommendation

For solving permutation or combination problems, you need to first visualize how the placements in positions for permutation, or selections for combination will actually be done for satisfying the given conditions. This is what we call problem modelling. After deciding clearly how the permutations or selections are to be carried out, the formulas will be used. Do not think in terms of formulas at first.

You are here