Permutations and combinations are fascinating branches of mathematics that deal with counting and arranging items. Whether you’re trying to figure out how many ways to arrange books on a shelf or selecting a team from a group, these concepts come into play. For Class 11 students, understanding these principles is essential, not just for exams but for various real-life applications. Dive into this set of carefully curated MCQs to test and enhance your grasp of permutations and combinations.

## 1. How many ways can 5 books be arranged on a shelf?

### Answer:

### Explanation:

It is a permutation of 5 items, which is 5P5 = 5! = 120.

## 2. How many ways can you select 2 books out of 5?

### Answer:

### Explanation:

It's a combination. 5C2 = 5! / (2!3!) = 10.

## 3. How many ways can 4 letters be arranged out of the word "MATH"?

### Answer:

### Explanation:

It's a permutation of 4 distinct items: 4! = 24.

## 4. In how many ways can 3 students be chosen from a class of 10 to represent their school?

### Answer:

### Explanation:

It's a combination. 10C3 = 10! / (3!7!) = 90.

## 5. The number of ways of choosing 2 cards from a pack of 52 cards is:

### Answer:

### Explanation:

52C2 = 52! / (2!50!) = 1326.

## 6. The number of permutations of n things taken all at a time, where p are of the same kind, q are of the same kind, and the rest are different is:

### Answer:

### Explanation:

Due to the repetition of similar items, we divide by their factorial.

## 7. In how many ways can 5 prizes be distributed among 4 students when each student can get any number of prizes?

### Answer:

### Explanation:

Each prize can be given in 4 ways. So, 4^5 = 625 ways.

## 8. How many 3-digit numbers can be formed using the digits 1, 2, 3, 4, and 5 without repetition?

### Answer:

### Explanation:

The first position has 5 choices, the second has 4, and the third has 3. So, 5×4×3 = 60.

## 9. The number of ways in which 8 distinct toys can be given to 3 children such that any child can get any number of toys is:

### Answer:

### Explanation:

Each toy can be given in 3 ways. So, 3^8 = 6561 ways.

## 10. If nPr = 5040 and nCr = 210, then the values of n and r are:

### Answer:

### Explanation:

From the given nPr, n!/(n-r)! = 5040, this hints towards n=9. Using nCr, we get 9C4 = 210.

## 11. How many words with or without meaning can be formed by using all the letters of the word 'TRIANGLE'?

### Answer:

### Explanation:

There are 8 distinct letters. Number of words = 8! = 40320.

## 12. The number of diagonals of a decagon is:

### Answer:

### Explanation:

Number of diagonals = n(n-3)/2 where n=10, so 10(10-3)/2 = 45.

## 13. The number of ways to select a committee of 3 people from a group of 6 people is:

### Answer:

### Explanation:

6C3 = 6! / (3!3!) = 20.

## 14. How many ways can 5 girls and 3 boys be seated in a row such that all the girls sit together?

### Answer:

### Explanation:

Treat the 5 girls as a single entity. So, there are 4 entities which can be arranged in 4!. The 5 girls among themselves can be arranged in 5!. Total = 4! * 5! = 2880.

## 15. In how many ways can the letters of the word 'COMBINE' be arranged?

### Answer:

### Explanation:

There are 7 distinct letters, so 7! = 5040.

## 16. The number of ways of selecting 3 consonants out of 7 and 2 vowels out of 4 is:

### Answer:

### Explanation:

Number of ways = 7C3 * 4C2 = 210 * 5 = 1050.

## 17. The number of ways of arranging n distinct things taken r at a time, when repetition is allowed, is:

### Answer:

### Explanation:

Since repetition is allowed, each position can be filled in n ways.

## 18. How many words can be formed using the letters of the word 'LETTER', such that E occurs together?

### Answer:

### Explanation:

Treat the two E's as a single entity. Then we arrange LETTER, which is 6 letters, but with T being repeated, so 5!/2!. The E's can be arranged among themselves in 2! ways, so total = 2! * 5!/2! = 60.

## 19. In how many ways can 4 boys and 4 girls be seated around a circular table such that boys and girls are alternate?

### Answer:

### Explanation:

Fix one boy at a position. Now, the 3 boys can be arranged in 3!, and the 4 girls in 4!, so 3! * 4! = 576.

## 20. If nC10 = nC9, then n is:

### Answer:

### Explanation:

For nC10 = nC9, n = 10 + 9 = 19.