Chapter 6: Permutations and Combinations

Ans: There will be however many ways as there are methods of filling 3 empty spots in progression by the given five digits. In this case, repetition of digits is allowed.

Therefore, the units spot can be filled in by any of the given five digits.

Similarity, tens and hundred digits can be filled in by any of the given five digits.

Hence, by the multiplication principle, the number of ways in which three-digit numbers can be formed from the given digits is $5 \times 5 \times 5 = 125$.

Ans: In this case, repetition of digits is not allowed. Here, on the off chance that unit place is filled in first, it very well may be filled by any of the given five digits.

Therefore, the number of ways of filling the units spot of the three-digit number is 5.

Then, at that point, the tens spot can be filled with any of the leftover four digits and the hundreds spot can be filled with any of the excess three digits. Thus, by the multiplication principle, the number of ways in which three-digit numbers can be formed without repeating the given digits is $5 \times 4 \times 3 = 60$.

Ans: There will be as many ways as there are ways of filling 3 vacant places in succession by the given six digits.

For this situation, the units spot can be filled by 2 or 4 or 6 just i.e., the units spot can be filled in 3 ways.

The tens spot can be filled by any of the 6 digits in 6 distinctive manners and furthermore the hundreds spot can be filled by any of the 6 digits in 6 diverse manners, as the digits can be repeated. Thus, by multiplication principle, the required number of three digit even numbers is $3 \times 6 \times 6 = 108$.

Ans: There are as many codes as there are ways of filling 4 vacant places in succession by the first 10 letters of the English alphabet, keeping in mind that the repetition of letters is not allowed.

The primary spot can be filled in 10 distinctive manners by any of the initial 10 letters of the English letters in order following which, the runner up can be filled in by any of the leftover letters in 9 diverse manners. The third spot can be filled in by any of the excess 8 letters in 8 diverse manners and the fourth spot can be filled in by any of the leftover 7 letters in 7 distinctive manners.

Therefore, by multiplication principle, the required numbers of ways in which 4 empty places can be filled is $10 \times 9 \times 8 \times 7 = 5040$.

Hence, 5040 four-letter codes can be shaped utilizing the initial 10 letters of the English letters in order, if no letter is repeated.

Ans: Since, it is given that the 5-digit telephone numbers always start with 67.

Therefore, there will be as many phone numbers as there are methods of filling 3 empty spots 6, 7, __, __, __ by the digits 0 – 9, remembering that the digits cannot be repeated.

In this way, the units spot can be filled in 8 diverse manners following which, the tens spot can be filled in by any of the leftover 7 digits in 7 distinctive manners, and the hundreds spot can be filled in by any of the excess 6 digits in 6 distinctive manners. Therefore, by multiplication principle, the required number of ways in which 5-digit telephone numbers can be constructed is $8 \times 7 \times 6 = 336$.

Ans: When a coin is tossed once, the number of outcomes is 2 (Head and tail) i.e., in each throw, the number of ways of showing a different face is 2.

Thus, by multiplication principle, the required number of possible outcomes is $2 \times 2 \times 2 = 8$.

Ans: Since, each signal requires the use of 2 flags.

There will be however many flags as there are methods of filling in 2 empty spots in progression by the given 5 flags of various colours. 

The upper empty spot can be filled in 5 distinctive manners by any of the 5 flags following which, the lower empty spot can be filled in 4 diverse manners by any of the leftover 4 unique flags.

Thus, by multiplication principle, the number of different signals that can be generated is $5 \times 4 = 20$.