Permutation & Combination Practice Questions & Solutions for SBI PO Prelims

SBI PO prelims 2023 exam is starting from 1st November which is tomorrow and the exam is expected to continue on 4th & 5th Nov respectively for all shifts. SBI PO is quite an important exam for which many aspirants prepare every year. Mentioned below is the SBI PO Prelims exam pattern below: 

SBI PO Prelims Exam Pattern

Permutation and Combination:

Factorial:

Let ‘n’ be a natural number, then n! = n × (n – 1) × (n – 2) × … × 1

Permutation:

Permutation is the arrangement of different number of things where we take into consideration the order or arrangement.

Combination:

The number of different groups or selections that can be formed by taking some or all items at a time, is called combination.

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Permutation & Combination Practice Questions & Solutions for SBI PO Prelims

Question 1: BCCI has to select a team of 6 batsmen and 5 bowlers to be sent for England tour, while Dhoni (Batsman) and Ashwin (Bowler) is always selected. If there are total 10 batsmen and 8 bowlers in the team, then find the total number of ways of selection.

A) 4410

B) 5680

C) 6250

D) 7240

E) None of these

Question 2: Find the number of words that can be formed using the letters of the word “IRIDIUM” so that vowels are always together.

A) 48

B) 96

C) 120

D) 240

E) 560

Question 3: A group of 15 members has male members and female members in the ratio of 3:2 respectively. 3 members are to be selected from this group. Find the number of ways the members can be selected such that it has odd number of female members.

A) 224 ways

B) 254 ways

C) 242 ways

D) 236 ways

E) None of these

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Question 4: A monetary policy committee of six members is to be selected from a group of seven RBI directors and six government executives. In how many ways will the committee having at least two RBI directors and at least two government executives be selected?

A) 1716

B) 1540

C) 1260

D) 1050

E) 700

Question 5: A cricket team of 11 players is to be selected from a group of 8 batsmen and 7 bowlers. In how many ways a team having at least 6 batsmen and at least 4 bowlers are selected?

A) 966

B) 762

C) 854

D) 868

E) None of these

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Question 6: In a car, 3 girls and 2 boys can be accommodated. In how many ways the selection can be made out of 5 boys and 5 girls?

A) 120 ways

B) 80 ways

C) 100 ways

D) 50 ways

E) None of these

Question 7: How many 3-letters word with or without meaning, can be formed out of the letters of the word, ‘COMPUTERS’, if repetition of letters is not allowed?

A) 504

B) 720

C) 336

D) 360

E) 126

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Question 8: Five coloured rings have to be arranged on two shelves with maximum three rings on one shelf. Find the total possible number of ways to arrange the rings.

A) 160

B) 180

C) 200

D) 240

E) 150

Question 9: A team of 7 players is to be selected from a group of 6 male players and 5 female players. In how many ways a team having at least 5 male players is selected?

A) 60

B) 65

C) 70

D) 75

E) None of these

Question 10: Five articles out of which 3 are cups and rest are glasses have to be arranged on a shelf. Find the number of ways of arrangement in which all three cups are not placed together.

A) 80 ways

B) 72 ways

C) 60 ways

D) 84 ways

E) 64 ways

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प्रश्न 1: बीसीसीआई को इंग्लैंड दौरे के लिए 6 बल्लेबाजों और 5 गेंदबाजों की एक टीम का चयन करना होगा, जबकि धोनी (बल्लेबाज) और अश्विन (गेंदबाज) हमेशा चुने जाते हैं। यदि टीम में कुल 10 बल्लेबाज और 8 गेंदबाज़ हैं, तो चयन के तरीकों की कुल संख्या बताइए?

A) 4410

B) 5680

C) 6250

D) 7240

E) None of these

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प्रश्न 2: शब्द “IRIDIUM” ” के अक्षरों का उपयोग करके गठित शब्दों की संख्या बताये जिसमें vowels हमेशा एक साथ रहें।

A) 48

B) 96

C) 120

D) 240

E) 560

प्रश्न 3: 15 सदस्यों के एक समूह में पुरुष सदस्य और महिला सदस्य का अनुपात क्रमशः 3: 2 हैं। इस समूह से 3 सदस्यों का चयन किया जाना है।ऐसे कितने तरीके से सदस्य का गठन किया जा सकता है कि इसमें महिला सदस्यों की विषम संख्या हो|

A) 224 ways

B) 254 ways

C) 242 ways

D) 236 ways

E) इनमें से कोई नहीं

प्रश्न 4: सात RBI directors और छह government executives के एक समूह से छह सदस्यों की मौद्रिक नीति समिति का चयन किया जाना है।ऐसे कितने प्रकार से समिति का गठन किया जा सकता है कि समिति में कम से कम दो RBI directors और कम से कम दो government executives का चयन हो?

A) 1716

B) 1540

C) 1260

D) 1050

E) 700

प्रश्न 5: 11 खिलाड़ियों की एक क्रिकेट टीम को 8 बल्लेबाजों और 7 गेंदबाजों के एक समूह से चयन किया जा रहा है।कितने प्रकार से टीम का चयन इस तरह किया जा सकता है कि टीम में कम से कम 6 बल्लेबाज और कम से कम 4 गेंदबाज हो?

A) 966

B) 762

C) 854

D) 868

E) इनमें से कोई नहीं

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प्रश्न 6: एक कार में 3 लड़कियों और 2 लड़कों को रखा जा सकता है। 5 लड़कों और 5 लड़कियों में से कितने तरीकों से चयन किया जा सकता है?

A) 120 ways

B) 80 ways

C) 100 ways

D) 50 ways

E) इनमें से कोई नहीं

प्रश्न 7: यदि अक्षरों की पुनरावृत्ति नहीं की जाए तो कितने 3-अक्षर के सार्थक या निरर्थक शब्द, ‘COMPUTERS’ शब्द से बनाए जा सकते हैं?

A) 504

B) 720

C) 336

D) 360

E) 126

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प्रश्न 8: पांच रंग के rings को दो shelves पर ऐसे व्यवस्थित किया जाना है कि एक shelf पर अधिकतम तीन rings हो| कुल कितने तरीकों से rings की व्यवस्था की जा सकती है ?

A) 160

B) 180

C) 200

D) 240

E) 150

प्रश्न 9: 7 खिलाड़ियों की एक टीम को 6 पुरुष खिलाड़ियों और 5 महिला खिलाड़ियों के समूह से चुना जाना है। कितने तरीकों से न्यूनतम 5 पुरुष खिलाड़ियों वाली टीम का चुनाव किया जा सकता है?

A) 60

B) 65

C) 70

D) 75

E) इनमें से कोई नहीं

प्रश्न 10: पाँच वस्तुएँ जिनमें से 3 कप हैं और शेष गिलास हैं, को एक शेल्फ पर व्यवस्थित करना है। व्यवस्था के उन तरीकों की संख्या ज्ञात कीजिए जिनमें तीनों कप एक साथ नहीं रखे गए हैं।

A) 80 ways

B) 72 ways

C) 60 ways

D) 84 ways

E) 64 ways

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ANSWER KEYS and SOLUTIONS:

Solution 1: A)

As Dhoni (Batsman) and Ashwin (Bowler) is always selected,

So, we have to select only (6 – 1) = 5 batsmen, out of (10 –1) = 9 batsmen, and (5 – 1) = 4 bowlers out of (8 – 1) = 7 bowlers.

So, required number of ways = 9C5 × 7C4 = 126 × 35 = 4410

Hence, option a.

Solution 2: B)

The word “IRIDIUM” contains 7 letters with 4 vowels, (3I, R, D, U, and M)

If vowels comes together, then 4 vowels will behave as a single entity, so, remaining 3 consonants and this entity, a total of 4 letters will arrange themselves in 4! Ways, and 4 vowels (3I and U) will arrange themselves in (4!/3!) Ways,

Number of words that can be formed using the letters of the word “IRIDIUM” so that vowels are always together =

(4!) × (4!/3!) = 24 × 4 = 96

Hence, option b.

Solution 3: D)

Number of male members = (3/5) × 15 = 9

Number of female members = 15 – 9 = 6

Required number of ways = 6C1 × 9C2 + 6C3 × 9C0

= 216 + 20 = 236 ways

Hence, option d.

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Solution 4: B)

Case I: 2 RBI directors and 4 govt. executives are there in the committee

Total number of ways = 7C2 × 6C4 = 21 × 15 = 315

Case II: 3 RBI directors and 3 govt. executives are there in the committee

Total number of ways = 7C3 × 6C3 = 35 × 20 = 700

Case III: 4 RBI directors and 2 govt. executives are there in the committee

Total number of ways = 7C4 × 6C2 = 35 × 15 = 525

So, total number of ways = 315 + 700 + 525 = 1540

Hence, option b.

Solution 5: D)

Case I: 6 batsmen and 5 bowlers are there in the team.

Number of ways = 8C6 × 7C5 = 28 × 21 = 588

Case II: 7 batsmen and 4 bowlers are there in the team.

Number of ways = 8C7 × 7C4 = 8 × 35 = 280

So, required total number of ways to select the team = 588 + 280 = 868

Hence, option d.

Solution 6: C)

Required number of ways = 5C2 × 5C3 = 10 × 10 = 100 ways

Hence, option c.

Solution 7: A)

Number of letters = 9

Required number of words = 9P3= 9 × 8 × 7 = 504

Hence, option a.

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Solution 8: D)

Number of ways to arrange three rings on one shelf = 5C3 × 3! = 60 ways

Number of ways to arrange two rings on second shelf = 2! = 2 ways

Number ofways to arrange the rings on either of the shelves = 2! = 2 ways

Total number of possible ways = 60 × 2 × 2 = 240

Hence, option d.

Solution 9: B)

Case I: 5 male players and 2 female players are there in the team

Number of ways = 6C5 × 5C2 = 6 × 10 = 60 ways

Case I: 6 male players and a female player are there in the team

Number of ways = 6C6 × 5C1 = 1 × 5 = 5 ways

Total number of ways = 60 + 5 = 65 ways

Hence, option b.

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Solution 10: D)

Since there are 5 articles, total number of ways of arrangement = 5! = 120 ways

If we consider 3 cups as one article, therefore, number of ways of arrangement of cups among themselves = 3! = 6 ways

Number of ways of arrangement of remaining three articles = 3! = 6 ways

Therefore, required number of ways = 120 – (6 × 6) = 84 ways

Hence, option d.

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