Selection and Combination Questions with Solutions — Set 10 (Q91 to Q94)

This set contains Selection and Combination Questions with Solutions — Set 10 (Q91-Q94) covering a mix of question types and difficulty levels — from basic to advanced — exactly as asked in real competitive exams.

Solutions are written in a simple, step-by-step notebook style for easy self-study and quick understanding. Each solution is broken down step by step so even the toughest question feels easy. These questions are hand-picked for students preparing for SSC CGL, SSC CHSL, CAT, Bank PO, Bank Clerk, UPSC CSAT, Railway RRB, AMCAT, eLitmus, TCS NQT and all campus placement aptitude tests. International students preparing for GRE, GMAT, SAT, ACT, MAT and all Numerical Reasoning Tests will find these equally useful.

✏️ Attempt each question on your own first — then check the solution below.

91. 7 relative of a man comprises 4 ladies and 3 gentlmen; his wife has also 7 relatives; 3 of them are ladies & 4 gentlemen. In how many ways can they invite a dinner party of 3 ladies and 3 gentlemen so that there are 3 of man’s relative and 3 of the wife’s relative?

92. A set S has n elements. what is the number of subsets of this set which contains odd number of elements?

93. 8 people are to be selected for a work from a group of 20 people such that exactly one from Mr. & Mrs. A is selected, similarly one from Mr. & Mrs. B and Mr. & Mrs. C and Mr. & Mrs. D is selected for the work. How many ways this can be done?

94. In an election 3 person are to be selected from 6 candidates. A voter can cast any number of votes but not more than the candidates to be elected. In how many ways can he cast his vote?

91. 7 relative of a man comprises 4 ladies and 3 gentlmen; his wife has also 7 relatives; 3 of them are ladies & 4 gentlemen. In how many ways can they invite a dinner party of 3 ladies and 3 gentlemen so that there are 3 of man’s relative and 3 of the wife’s relative?
Solution:-
This question can be solved by making different cases given below:-

Hence total number of ways = 16 + 1 + 144 + 324
                                                   = 485 ways         Answer

92. A set S has n elements. what is the number of subsets of this set which contains odd number of elements?
Solution:-
Total number of subset of this set = 2ⁿ
out of these half is even & half contains odd elements.
Hence required number =  \(\\boldsymbol{frac{{{2^n}}}{2}}\)  = 2ⁿ⁻¹            Answer

93. 8 people are to be selected for a work from a group of 20 people such that exactly one from Mr. & Mrs. A is selected, similarly one from Mr. & Mrs. B and Mr. & Mrs. C and Mr. & Mrs. D is selected for the work. How many ways this can be done?
Solution:-
4 people from the group of Mr. & Mrs. A, B, C, D can be selected in 2×2×2×2 = 16 ways.
Now from remaining 20 – 2×4 = 12 people, 4 peoples can be selected in ¹²C₄ = 495 ways
∴ Total number of ways for selection
= 16×495
= 7920 ways      Answer

94. In an election 3 person are to be selected from 6 candidates. A voter can cast any number of votes but not more than the candidates to be elected. In how many ways can he cast his vote?
Solution:-
A voter can vote for either 1 or 2 or 3 contestants.
Hence required number of ways is
= ⁶C₁ + ⁶C₂ + ⁶C₃
= 6 + 15 + 20
= 41 ways             Answer

✅ Well done on completing Set 10!

Now revisit the Selection and Combination concept page to strengthen your formulas and tricks before moving ahead.

Consistent practice is the key to mastering percentage for SSC CGL, SSC CHSL, CAT, Bank PO, Bank Clerk, UPSC CSAT, Railway RRB, AMCAT, eLitmus, TCS NQT and international exams including GRE, GMAT, SAT, ACT, MAT and all Numerical Reasoning Tests. Want to understand the concept better? Read about Selection and Combination Wikipedia before attempting the next set.

This page is part of our complete series of percentage questions with solutions for competitive exams — covering every question type from basic to advanced so you can build speed, accuracy and confidence. Practising these questions regularly will also strengthen your core percentage concept before your exam day.