How many 5-digit odd numbers can be formed using the digits 3, 4, 5, 6, 7, 8, and 9 if:
i) repetition of digits is not allowed?
ii) repetition of digits is allowed?

How many ways can four boys and three girls be seated in a row containing seven seats if:
a) they may sit anywhere?
b) all three girls are together?
c) the boys and girls must alternate?
d) the girls are to occupy odd seats?

Calculate the number of 5-digit odd numbers without repetition: 6 × 5 × 4 × 3 × 4 = 1440 .
Calculate the number of 5-digit odd numbers with repetition: 7 4 × 4 = 9604 .
Calculate the number of ways to seat 4 boys and 3 girls without restrictions: 7 ! = 5040 .
Calculate the number of ways to seat 4 boys and 3 girls with the girls together: 5 ! × 3 ! = 720 .
Calculate the number of ways to seat 4 boys and 3 girls alternating: 4 ! × 3 ! = 144 .
Calculate the number of ways to seat 4 boys and 3 girls with girls in odd seats: ( 3 4 ​ ) × 3 ! × 4 ! = 576 .
1440 , 9604 , 5040 , 720 , 144 , 576 ​

Explanation

Problem Analysis We are given the digits 3, 4, 5, 6, 7, 8, and 9. We need to find the number of 5-digit odd numbers that can be formed using these digits, with and without repetition. We also need to find the number of ways to seat 4 boys and 3 girls in a row of 7 seats under different conditions.

5-Digit Odd Numbers Without Repetition For part (i), we want to find the number of 5-digit odd numbers without repetition. The last digit must be odd, so we have 4 choices (3, 5, 7, 9). After choosing the last digit, we have 6 remaining digits for the first position, 5 for the second, 4 for the third, and 3 for the fourth. Therefore, the total number of such numbers is calculated as follows: 6 × 5 × 4 × 3 × 4 = 1440

5-Digit Odd Numbers With Repetition For part (ii), we want to find the number of 5-digit odd numbers with repetition allowed. The last digit must be odd, so we have 4 choices (3, 5, 7, 9). Since repetition is allowed, we have 7 choices for each of the other four digits. Therefore, the total number of such numbers is calculated as follows: 7 × 7 × 7 × 7 × 4 = 7 4 × 4 = 2401 × 4 = 9604

Seating Arrangement Without Restrictions For part (a), we want to find the number of ways to seat 4 boys and 3 girls in a row of 7 seats without any restrictions. This is simply arranging 7 people in 7 seats, which can be done in 7 ! ways. Therefore, the total number of arrangements is calculated as follows: 7 ! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040

Seating Arrangement With Girls Together For part (b), we want to find the number of ways to seat 4 boys and 3 girls such that all three girls are together. We can consider the three girls as a single unit. So, we have 4 boys and 1 unit of girls, which makes a total of 5 entities to arrange. These 5 entities can be arranged in 5 ! ways. The 3 girls within their unit can be arranged in 3 ! ways. Therefore, the total number of arrangements is calculated as follows: 5 ! × 3 ! = ( 5 × 4 × 3 × 2 × 1 ) × ( 3 × 2 × 1 ) = 120 × 6 = 720

Seating Arrangement With Alternating Boys and Girls For part (c), we want to find the number of ways to seat 4 boys and 3 girls such that the boys and girls must alternate. Since there are 4 boys and 3 girls, the arrangement must be BGBGBGB. The 4 boys can be arranged in 4 ! ways, and the 3 girls can be arranged in 3 ! ways. Therefore, the total number of arrangements is calculated as follows: 4 ! × 3 ! = ( 4 × 3 × 2 × 1 ) × ( 3 × 2 × 1 ) = 24 × 6 = 144

Seating Arrangement With Girls in Odd Seats For part (d), we want to find the number of ways to seat 4 boys and 3 girls such that the girls occupy odd seats. There are 4 odd seats (1st, 3rd, 5th, 7th). We need to choose 3 of these 4 seats for the girls, which can be done in ( 3 4 ​ ) ways. The 3 girls can be arranged in these 3 seats in 3 ! ways. The 4 boys can be arranged in the remaining 4 even seats in 4 ! ways. Therefore, the total number of arrangements is calculated as follows: ( 3 4 ​ ) × 3 ! × 4 ! = 4 × ( 3 × 2 × 1 ) × ( 4 × 3 × 2 × 1 ) = 4 × 6 × 24 = 576

Final Answer Therefore, the answers are: i) 1440 ii) 9604 a) 5040 b) 720 c) 144 d) 576


Examples
These types of counting problems are useful in many real-world scenarios. For example, calculating the number of possible passwords that meet certain criteria (like having a certain number of digits and special characters) is a similar problem to counting the number of 5-digit odd numbers. Seating arrangement problems are applicable in event planning, where you might need to arrange guests with certain constraints, such as keeping certain people together or alternating groups.

Answered by GinnyAnswer | 2025-07-03

There are 1440 five-digit odd numbers possible without repetition and 9604 with repetition. When seating four boys and three girls, 5040 arrangements are possible without restrictions, whereas specific constraints yield 720 for girls together, 144 for alternating, and 576 for girls in odd seats.
;

Answered by Anonymous | 2025-07-04