CBSE Class 11 Mathematics Chapter 7: Permutations and Combinations NCERT Solutions

We need to form 3-digit numbers using the digits {1, 2, 3, 4, 5}.

1. Repetition of digits is allowed:

   For a 3-digit number, there are three places to fill: units, tens, and hundreds. Since repetition is allowed, each of these places can be filled by any of the 5 given digits.

   • Number of choices for the units place = 5
   • Number of choices for the tens place = 5
   • Number of choices for the hundreds place = 5

   By the multiplication principle, the total number of 3-digit numbers that can be formed is the product of the number of choices for each place.

   $$Total\; numbers\; =\; 5\; \backslash times\; 5\; \backslash times\; 5\; =\; 125$$

   Thus, 125 three-digit numbers can be formed if repetition is allowed.

2. Repetition of digits is not allowed:

   In this case, once a digit is used for a place, it cannot be used again for another place.

   • Number of choices for the units place = 5 (any of the 5 digits)
   • After filling the units place, there are 4 remaining digits. So, number of choices for the tens place = 4.
   • After filling the units and tens places, there are 3 remaining digits. So, number of choices for the hundreds place = 3.

   By the multiplication principle, the total number of 3-digit numbers that can be formed without repetition is:

   $$Total\; numbers\; =\; 5\; \backslash times\; 4\; \backslash times\; 3\; =\; 60$$

   Thus, 60 three-digit numbers can be formed if repetition is not allowed.

We need to form 3-digit even numbers using the digits {1, 2, 3, 4, 5, 6}, with repetition allowed.

For a number to be even, its units digit must be an even number. From the given digits, the even digits are {2, 4, 6}.

• Number of choices for the units place = 3 (i.e., 2, 4, or 6)
• Since repetition is allowed, the tens place can be filled by any of the 6 given digits. Number of choices for the tens place = 6.
• Similarly, the hundreds place can be filled by any of the 6 given digits. Number of choices for the hundreds place = 6.

By the multiplication principle, the total number of 3-digit even numbers that can be formed is:

$$Total\; numbers\; =\; 3\; \backslash times\; 6\; \backslash times\; 6\; =\; 108$$

Therefore, 108 three-digit even numbers can be formed.

We need to form a 4-letter code using the first 10 letters of the English alphabet (A to J), and no letter can be repeated. This is a permutation problem where we are arranging 4 letters out of 10.

There are 4 positions in the code to fill.

• For the first position, there are 10 choices (any of the first 10 letters).
• Since no letter can be repeated, for the second position, there are 9 remaining choices.
• For the third position, there are 8 remaining choices.
• For the fourth position, there are 7 remaining choices.

By the multiplication principle, the total number of 4-letter codes that can be formed is:

$$Total\; codes\; =\; 10\; \backslash times\; 9\; \backslash times\; 8\; \backslash times\; 7\; =\; 5040$$

Alternatively, this can be calculated using the permutation formula P(n, r) = $\backslash frac\{n!\}\{(n-r)!\}$, where n=10 and r=4.

$$P(10,\; 4)\; =\; \backslash frac\{10!\}\{(10-4)!\}\; =\; \backslash frac\{10!\}\{6!\}\; =\; 10\; \backslash times\; 9\; \backslash times\; 8\; \backslash times\; 7\; =\; 5040$$

Hence, 5040 four-letter codes can be formed.

We need to construct 5-digit telephone numbers using digits 0 to 9. The condition is that each number must start with '67' and no digit can appear more than once.

A 5-digit telephone number has 5 places. The first two digits are fixed as 6 and 7.

The number looks like: 6 7 _ _ _

The digits available are {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. Since the digits 6 and 7 have already been used and no digit can appear more than once, we have 8 remaining digits {0, 1, 2, 3, 4, 5, 8, 9} to fill the remaining 3 places.

• The third digit (from the left) can be any of the remaining 8 digits. Number of choices = 8.
• After filling the third digit, there are 7 digits left. So, the fourth digit can be any of the remaining 7 digits. Number of choices = 7.
• After filling the third and fourth digits, there are 6 digits left. So, the fifth digit can be any of the remaining 6 digits. Number of choices = 6.

By the multiplication principle, the total number of 5-digit telephone numbers that can be constructed is:

$$Total\; numbers\; =\; 1\; \backslash times\; 1\; \backslash times\; 8\; \backslash times\; 7\; \backslash times\; 6\; =\; 336$$

(Here, the first '1' represents the choice for the first digit '6', and the second '1' represents the choice for the second digit '7').

Therefore, 336 such 5-digit telephone numbers can be constructed.