# Permutations and Combinations - Class 11 - Mathematics

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## Extra Questions - Permutations and Combinations | NCERT | Mathematics | Class 11

The total number of 6-digit numbers that can be made using digits 1, 2, 3, 4, if all the digits should appear in the number at least once is:

A) 1560

B) 840

C) 1080

D) 480

The correct answer is **A) 1560**. The computation involves considering all possible distributions of the digits 1, 2, 3, and 4, where each digit is used at least once in forming a 6-digit number.

### Types of Number Arrangements

**One digit appears three times**: Take an example such as $1, 2, 3, 4, 4, 4$. The digit '4' is repeated three times. To calculate:**Selection of the thrice-appearing digit**: There are $4$ options (i.e., $\binom{4}{1} = 4$ ways).**Arranging these digits**: There are 6 slots, with one digit repeating three times. Arrange by $\frac{6!}{3!}$ (handle permutation with repetition).

Thus, the number of such combinations is: $$ \left(\frac{6!}{3!}\right) \times \binom{4}{1} = 480 $$

**Two digits appear twice each**: An arrangement example is $1, 2, 3, 3, 4, 4$. Here, both '3' and '4' are repeated twice.**Selection of the two digits that repeat**: $\binom{4}{2} = 6$ ways.**Arranging these digits**: There are 6 slots, with two pairs of digits repeating twice. Arrange by $\frac{6!}{2!2!}$.

Hence, the number of such arrangements is: $$ \left(\frac{6!}{2!2!}\right) \times \binom{4}{2} = 1080 $$

### Total Number of Valid Numbers

Combining both scenarios, the total number of valid six-digit numbers is: $$ 480 + 1080 = 1560 $$

Thus, the total number of **6-digit numbers** that can be made using the digits 1, 2, 3, 4, with each appearing at least once, is **1560** (Option A).

The number of permutations that can be made out of the letters of the word "EQUATION" which start with a consonant and end with a consonant is

(A) $2!6!$

(B) $3!6$!

(C) $3!5$!

(D) $2!5!$

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There are 10 professors and 20 students, out of whom a committee of 2 professors and 3 students is to be formed. Find the number of ways in which this can be done. Further, find in how many of these committees: (i) A particular professor is included. (ii) A particular student is included. (iii) A particular student is excluded.

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A student is to answer 10 out of 13 questions in an examination such that he must choose at least four from the first five questions. The number of choices available to him is

A) 140

B) 196

C) $\mathbf{280}$

D) 346

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The number of ways so that the birthdays of 6 people fall in exactly 3 calendar months is:

Option 1) 118800

Option 2) 36960

Option 3) 158400

Option 4) 47520