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}. We will consider two cases based on whether repetition is allowed.

1. Repetition of digits is allowed:

   We have 3 positions to fill: the hundreds place, the tens place, and the units place. Since repetition is allowed, for each of these three places, we can choose any of the 5 given digits.

   Using the fundamental principle of counting (multiplication principle), the total number of 3-digit numbers that can be formed is the product of the number of choices for each place.

   Number of choices for the hundreds place = 5 (any digit from 1 to 5)

   Number of choices for the tens place = 5 (any digit from 1 to 5)

   Number of choices for the units place = 5 (any digit from 1 to 5)

   Total number of 3-digit numbers = 5 \times 5 \times 5 = 125.

2. Repetition of digits is not allowed:

   We have 3 positions to fill: the hundreds place, the tens place, and the units place. Since repetition is not allowed, the choice for each subsequent place depends on the digits already chosen.

   Number of choices for the units place = 5 (any digit from 1 to 5).

   After filling the units place, there are 4 digits remaining.

   Number of choices for the tens place = 4 (any of the remaining 4 digits).

   After filling the units and tens places, there are 3 digits remaining.

   Number of choices for the hundreds place = 3 (any of the remaining 3 digits).

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

   Total number of 3-digit numbers = 5 \times 4 \times 3 = 60.

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

Let the 3-digit number be represented by three places: Hundreds, Tens, Units.

Units place: For the number to be even, the units digit must be an even number from the given set. The even digits are {2, 4, 6}. So, there are 3 choices for the units place.

Tens place: Since repetition of digits is allowed, any of the 6 given digits can be used for the tens place. So, there are 6 choices for the tens place.

Hundreds place: Similarly, since repetition is allowed, any of the 6 given digits can be used for the hundreds place. So, there are 6 choices for the hundreds place.

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

Total numbers = (Choices for Hundreds place) × (Choices for Tens place) × (Choices for Units place)

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

We need to form a 4-letter code using the first 10 letters of the English alphabet (A, B, C, D, E, F, G, H, I, J), with the condition that no letter can be repeated. This is a permutation problem where we are arranging 4 letters out of 10 distinct letters.

Let the 4 positions in the code be represented by four vacant places.

First place: We can choose any of the 10 letters. So, there are 10 choices.

Second place: Since no letter can be repeated, we have used one letter for the first place. Thus, there are 9 remaining letters to choose from. So, there are 9 choices.

Third place: We have used two letters for the first two places. Thus, there are 8 remaining letters to choose from. So, there are 8 choices.

Fourth place: We have used three letters for the first three places. Thus, there are 7 remaining letters to choose from. So, there are 7 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) = $\backslash frac\{10!\}\{(10-4)!\}\; =\; \backslash frac\{10!\}\{6!\}\; =\; 10\; \backslash times\; 9\; \backslash times\; 8\; \backslash times\; 7\; =\; 5040$.

We need to construct 5-digit telephone numbers using digits {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. The conditions are that the number must start with '67' and no digit can appear more than once.

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

First digit: Fixed as 6 (1 choice).

Second digit: Fixed as 7 (1 choice).

Third digit: We have used the digits 6 and 7. There are 10 total digits (0-9). So, there are $10\; -\; 2\; =\; 8$ remaining digits available for the third position. Thus, there are 8 choices.

Fourth digit: We have used three distinct digits (6, 7, and the digit chosen for the third place). So, there are $10\; -\; 3\; =\; 7$ remaining digits available for the fourth position. Thus, there are 7 choices.

Fifth digit: We have used four distinct digits. So, there are $10\; -\; 4\; =\; 6$ remaining digits available for the fifth position. Thus, there are 6 choices.

By the multiplication principle, the total number of such 5-digit telephone numbers is:

Total numbers = (Choices for 1st digit) × (Choices for 2nd digit) × (Choices for 3rd digit) × (Choices for 4th digit) × (Choices for 5th digit)

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