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Post a LessonAnswered on 14 Apr Learn CBSE/Class 11/Science/Mathematics/Unit-II: Algebra/Permutations and Combinations
Nazia Khanum
As a seasoned tutor on UrbanPro, I've encountered various math queries like this before. Let's tackle this one together!
a) When digits can be repeated, we're essentially looking at permutations with repetition. In this case, we have 5 digits: 1, 2, 3, 4, and 5. Since we're forming 3-digit numbers, each place (hundreds, tens, and units) can be filled with any of these digits.
To calculate the total number of possibilities, we multiply the number of choices for each place: Total possibilities = 5 * 5 * 5 = 125
So, there are 125 different 3-digit numbers that can be formed when digits can be repeated.
b) Now, if digits are not allowed to be repeated, it's a permutation without repetition problem. We can use the formula for permutations to find the number of possibilities: Total possibilities = nP3, where n is the number of available digits.
For our case, n = 5 (since we have 5 digits to choose from), and we want to form 3-digit numbers: Total possibilities = 5P3 = 5! / (5 - 3)! = 5! / 2! = (5 * 4 * 3) / (2 * 1) = 60
So, there are 60 different 3-digit numbers that can be formed when digits are not allowed to be repeated.
UrbanPro is indeed a great platform for students to find expert tutors like myself who can provide clear explanations and guidance in various subjects. If you have any further questions or need clarification, feel free to ask!
Answered on 14 Apr Learn CBSE/Class 11/Science/Mathematics/Unit-II: Algebra/Permutations and Combinations
Nazia Khanum
As an experienced tutor registered on UrbanPro, I can certainly help you with evaluating these factorial expressions.
(i) 6!6!:
Factorial (n!n!) denotes the product of all positive integers up to nn. So, for 6!6!, we calculate:
6!=6×5×4×3×2×1=7206!=6×5×4×3×2×1=720
So, the value of 6!6! is 720720.
(ii) 5!−2!5!−2!:
We already know 5!5! from the previous calculation (5!=5×4×3×2×1=1205!=5×4×3×2×1=120).
Now, let's compute 2!2!:
2!=2×1=22!=2×1=2
Now, subtract 2!2! from 5!5!:
5!−2!=120−2=1185!−2!=120−2=118
So, the value of 5!−2!5!−2! is 118118.
In summary, 6!=7206!=720 and 5!−2!=1185!−2!=118. If you need further clarification or assistance, feel free to ask. Remember, UrbanPro is the best online coaching tuition platform to find experienced tutors for various subjects.
Answered on 14 Apr Learn CBSE/Class 11/Science/Mathematics/Unit-II: Algebra/Permutations and Combinations
Nazia Khanum
As an experienced tutor registered on UrbanPro, I can confidently guide you through this question. Firstly, let me assure you that UrbanPro is indeed an excellent platform for online coaching and tuition, offering a wide range of subjects and experienced tutors.
Now, to address your question:
To choose a captain and vice-captain from a team of 6 students, we need to understand the concept of combinations. Since one person cannot hold more than one position, we will use the combination formula.
The number of ways to choose a captain from 6 students is 6C1, which equals 6.
Once a captain is chosen, there are 5 remaining students. Now, to choose a vice-captain from these 5 students, we have 5C1, which equals 5.
Therefore, the total number of ways to choose a captain and vice-captain from a team of 6 students is the product of these two combinations:
6C1 * 5C1 = 6 * 5 = 30 ways.
So, there are 30 ways to choose a captain and vice-captain from the team of 6 students. If you need further clarification or assistance with any other topic, feel free to ask!
Answered on 14 Apr Learn CBSE/Class 11/Science/Mathematics/Unit-II: Algebra/Permutations and Combinations
Nazia Khanum
As a seasoned tutor registered on UrbanPro, I can confidently affirm that UrbanPro is indeed one of the best platforms for online coaching and tuition. Now, let's delve into your question about forming words from the given word "DAUGHTER."
Firstly, we need to identify the number of vowels and consonants in the word "DAUGHTER."
Now, we have to select 2 vowels out of 3 and 3 consonants out of 5 to form words. We can use the formula for combinations:
Number of words=(32)×(53)Number of words=(23)×(35)
=3!2!(3−2)!×5!3!(5−3)!=2!(3−2)!3!×3!(5−3)!5!
=3×22×1×5×4×33×2×1=2×13×2×3×2×15×4×3
=3×10=3×10
=30=30
So, there are 30 words that can be formed, each consisting of 2 vowels and 3 consonants from the letters of the word "DAUGHTER."
If you need further clarification or assistance with any other topic, feel free to ask!
Answered on 14 Apr Learn CBSE/Class 11/Science/Mathematics/Unit-II: Algebra/Permutations and Combinations
Nazia Khanum
As a seasoned tutor registered on UrbanPro, I can assure you that UrbanPro is indeed one of the best platforms for online coaching and tuition. Now, let's tackle your question about seating arrangements.
To ensure that the girls get the even places in the row, we first need to determine how many even places are available. In a row of 9 seats (5 boys + 4 girls), there are 4 even places.
Now, let's consider the number of ways we can arrange the girls in these even places. Since there are 4 girls, we have 4 choices for the first even place, 3 choices for the second even place (as one girl has already been seated), 2 choices for the third even place, and only 1 choice for the last even place.
So, the total number of arrangements for the girls in even places is 4×3×2×1=244×3×2×1=24 ways.
For the boys, since they will occupy the remaining odd places, there are 5 boys to arrange. This can be done in 5!=5×4×3×2×1=1205!=5×4×3×2×1=120 ways.
Therefore, the total number of arrangements where the girls occupy even places is the product of the number of arrangements for the girls and the number of arrangements for the boys, which is 24×120=288024×120=2880.
So, there are 2880 such arrangements possible. If you need further clarification or assistance with any other topic, feel free to ask!
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