Permutations And Combinations

To find the total number of positive integral solutions for the equation , we need to explore the factors of 24 with three positive integer variables.

First, perform the prime factorization of 24:

24 can be expressed as

.

Now, we need to distribute these prime factors among , , and .

Using a stars and bars approach (distribution of indistinguishable objects into distinguishable boxes):

1. Distribute : We distribute 3 objects (the three 2's) into 3 distinct boxes (x, y, and z):

The formula for this is , where n is the number of objects, and k is the number of boxes. For and ,

.

1. Distribute : We distribute 1 object (the one 3) into 3 distinct boxes:

.

Thus, the total number of ways to assign these factors to x, y, z is the product of the outcomes of these distributions:

.

Therefore, the total number of positive integral solutions for the equation is 30.

The correct answer is (D) : 430
By multinomial theorem, no. of ways to distribute 30 identical candies among four children C₁, C₂ and C₃, C₄
= Coefficient of x³⁰ in (x⁴ + x⁵ + … + x⁷) (x² + x³ +…+ x⁶) (1 + x + x²…)²
=Coefficient of x²⁴ in
= Coefficient of x²⁴ in (1 – x⁴ – x⁵ + x⁹) (1 – x)^–4
= ²⁷C₂₄ – ²³C₂₀ – ²²C₁₉ + ¹⁸C₁₅ = 430

The correct answer is 30
Number must start by 1 or 2 and for divisibility by
4 last two digits shall be divisible by 4

⇒ Total 30 numbers

The correct answer is (A) : 1411

Sum of all given numbers = 31

Difference between odd and even positions must be 0, 11 or 22, but 0 and 22 are not possible.
Therefore , Only difference 11 is possible
This is possible only when either 1, 2, 3, 4 is filled in odd position in some order and remaining in other
order. Similar arrangements of 2, 3, 5 or 7, 2, 1 or 4, 5, 1 at even positions.
∴ Total possible arrangements = (4! × 3!) × 4
= 576

The correct answer is 102




∴ m = 61 , n = 41
Therefore , m + n = 102

The correct answer is 1120
Required number of ways = Total ways of selection – ways in which B₁ and B₂ are present together.

=10(120−8)
= 1120

The correct answer is 1086
Let pqrs is 4-digit number.
So , first 3-digit pqr should be divisible by 's'
If s=1 , then numbers = 9×10×10 = 900
If s=2 , then numbers = 4×5×5 = 100
If s=3, then numbers = 3×4×4 = 48
If s=4, then numbers = 2×3×3 = 18
If s=5, then numbers = 1×2×2 = 4 numbers
If s=6,7,8 and 9 then , numbers = 4×4 = 16
∴ Total 4 digit Number = 900 + 100 + 48 + 18 + 4 + 16 = 1086

Given:

We are tasked to determine the rank of the word "THAMS" in alphabetical order using the factorial method.

Step 1: Assign Numerical Positions to Letters

The alphabetical order of the letters T, H, A, M, S is:

Thus, the word "THAMS" corresponds to the sequence: .

Step 2: Count Permutations for Each Letter

• For : There are 4 letters after in alphabetical order ( ). The number of permutations is:
• For : After , there is 1 letter ( ) that comes earlier in the sequence. The number of permutations is:
• For : There are no letters after that need to be considered. The number of permutations is:
• For : Similarly, there are no letters after to consider. The number of permutations is:
• For : There are no letters after . The number of permutations is:

Step 3: Compute Total Permutations Before "THAMS"

Total permutations before "THAMS" is given by:

Step 4: Add 1 for the Word Itself

Rank of "THAMS" is:

Final Answer:

The rank of the word "THAMS" is 103.

1) Letter multiset

MATHEMATICS (11 letters):
.

2) Total arrangements (no restriction)

3) Arrangements with C and S together

Treat the block (or ) as one item. Then we have 10 items: the block .
External permutations: Hence

4) Required count (not together)

5) Express as

Write . Then Therefore .

Final: Number of required words . Hence, .

Step 1: Identify the digits and their properties
The digits available are 1, 3, 5, and 8. We need to check which combinations of these digits can yield a sum that is divisible by 3, as a number is divisible by 3 if the sum of its digits is divisible by 3.

Step 2: Calculate the sum of the digits
The possible sums of three digits can range from:
- Minimum sum: 1 + 1 + 1 = 3
- Maximum sum: 8 + 8 + 8 = 24

Step 3: Identify the sums that are divisible by 3
We need to find the sums that are divisible by 3 within the range of 3 to 24. The sums that are divisible by 3 are:
- 3, 6, 9, 12, 15, 18, 21, 24

Step 4: Determine combinations for each valid sum
Now, we will analyze the combinations of digits that can yield each of these sums:

1. Sum = 3: (1, 1, 1)
2. Sum = 6: (1, 1, 3) or (1, 3, 1) or (3, 1, 1) → 3 combinations
3. Sum = 9: (1, 1, 5), (1, 3, 3), (3, 1, 5), (3, 3, 3), (5, 1, 3), (5, 3, 1) → 6 combinations
4. Sum = 12: (1, 5, 5), (3, 3, 5), (3, 5, 3), (5, 3, 3), (5, 5, 1) → 5 combinations
5. Sum = 15: (5, 5, 5) → 1 combination
6. Sum = 18: (5, 5, 8), (5, 8, 5), (8, 5, 5) → 3 combinations
7. Sum = 21: (5, 8, 8), (8, 5, 8), (8, 8, 5) → 3 combinations
8. Sum = 24: (8, 8, 8) → 1 combination

Step 5: Count the total combinations
Now we will count the total combinations for each valid sum:
- Sum = 3: 1 combination
- Sum = 6: 3 combinations
- Sum = 9: 6 combinations
- Sum = 12: 5 combinations
- Sum = 15: 1 combination
- Sum = 18: 3 combinations
- Sum = 21: 3 combinations
- Sum = 24: 1 combination

Adding these together:
1 + 3 + 6 + 5 + 1 + 3 + 3 + 1 = 23 combinations

Step 6: Calculate the total number of three-digit numbers
Since each combination can be arranged differently, we need to calculate the permutations for each case.

For example:
- For combinations with all distinct digits, the number of arrangements is 3! = 6.
- For combinations with two identical digits, the number of arrangements is 3!/2! = 3.
- For combinations with all identical digits, the number of arrangements is 1.

Final Count
After calculating the arrangements for each valid combination, we sum them up to get the total number of three-digit numbers divisible by 3.

Conclusion
The total number of three-digit numbers divisible by 3 that can be formed using the digits 1, 3, 5, and 8 (with repetition allowed) is 23.

The correct answer is 432
Vowels - I, U, E,
Consonants - N, V, R, S
⇒ ³C₂ × ⁴C₂ × 4!
= 3 × 6 × 24
= 432
So, 432 four letters word can be made

The rank of the word "PUBLIC" is:
= (4x5! + 4x4! + 0x3! + 2x2! + 1x1! + 0x0!) + 1
= ((4x120) + 96 + 0 + 4 + 1 + 0) + 1
= 480 + 100 + 2
= 582

The correct answer is 44
Number of seating arrangements

A five-digit number is divisible by 5 if its last digit is either 0 or 5. The number must also be greater than 40000.

Case 1: The last digit is 0 - If the last digit is 0, the first digit can be 5, 7, or 9 (since the number must be greater than 40000). This gives us 3 choices for the first digit.
- For the remaining three digits, we have 4 remaining choices (we’ve used two digits already), and these can be arranged in:
- So, the number of five-digit numbers ending in 0 is:

Case 2: The last digit is 5 - If the last digit is 5, the first digit can be 7 or 9 (since the number must be greater than 40000, and we can’t use 5 again). This gives us 2 choices for the first digit.
- For the remaining three digits, we have 4 remaining choices, and they can be arranged in:
- So, the number of five-digit numbers ending in 5 is:

Total Number of Five-Digit Numbers: Adding the counts from both cases, we get:

Final Answer: There are 120 such five-digit numbers.

The correct answer is 327.

Rank =
= 240+72+12+2+1
= 327

Step 1: Understanding the problem There are 8 persons to be transported, and there are 3 cars. Each car can carry at most 3 persons. We need to calculate the number of ways to assign these 8 persons to the cars.
Step 2: Distribute the persons in the cars To distribute the 8 persons into 3 cars, with each car holding a maximum of 3 persons, we first assign 3 persons to two cars and 2 persons to the third car. The number of ways to select 3 persons for the first car from the 8 is given by: Now, 5 persons remain, so the number of ways to select 3 persons for the second car is: Finally, 2 persons remain, so the number of ways to assign them to the third car is:
Step 3: Consider the arrangements of the cars Since there are 3 cars of different makes, the arrangement of the cars is important. Therefore, the number of ways to assign the selected persons to the cars is multiplied by the number of ways to arrange the cars, which is 3! (since there are 3 cars). Thus, the number of ways in which the persons can be transported is 1680.

The number must be divisible by , so it must be divisible by both and . For divisibility by , the last digit must be . For divisibility by , the sum of the digits must be divisible by .
Let us consider different cases based on the distribution of digits:
Case 1: All digits are the same
For , there is only number.
Case 2: Two distinct digits
: Numbers of the form :

: Numbers of the form :

Case 3: Three distinct digits
For , let us consider the permutations:
Digits: :

Digits: :

Digits: :

Digits: :

Total

Conclusion
The total number of such numbers is:

Let the number of couples be .

The total number of ways is given by:

.

Step 1: Simplify the equation:

.

.

Step 2: Factorize :

.

Thus, .

Step 3: Calculate the total number of persons:

Total persons = .

Final Answer: The number of persons is .

The numbers are .

• Step 1: Numbers having the string :
  • Positions:
  • Total permutations:
• Step 2: Numbers having the string :
  • Positions:
  • Total permutations:
• Step 3: Numbers having both strings and :
  • Positions:
  • Total permutations:
• Step 4: Apply the Inclusion-Exclusion Principle:

.

• Simplify:

.

• Step 5: Total numbers:

Total permutations: .

• Step 6: Numbers having neither nor :

Required numbers = .

Final Answer: The required numbers are .

We are given:

• (1)
• (2)

Subtract equation (1) from equation (2):

Testing integral values of :

For , we have:

Thus, . Substituting back into equation (1) to find :

Now, compute :

Find the positive integer divisors of 38:

• 1, 2, 19, 38

Sum of divisors:

Thus, the correct answer is option (2).

The correct answer is : 31

3 sis prime number,

22+7+2+0+………

= 31

maximum value of n=31

Total Cases:

Subtracting Invalid Cases:

• Case 1: One child receives no orange:
• Case 2: Two children receive no orange:

Final Calculation:

To solve this problem, we need to determine whether the expressions and belong to the set of natural numbers .

1. Expression for :
   We have . First, calculate : .
   Now substitute into the expression for : .
   Here, represents the factorial of , which is the product of all positive integers up to .
   The denominator, , suggests raising to the power .
   Since is a factorial, it includes as a factor many times, ensuring that will divide without leaving any decimal or fraction, thus .
2. Expression for :
   We have . Calculate : .
   Now substitute into the expression for : .
   Again, includes as a factor repeatedly, more than sufficient for ensuring divides completely, so .
3. Conclusion:
   Both and are in .

To find the number of ways to distribute 21 identical apples among three children such that each child gets at least 2 apples, we can solve the problem using the "stars and bars" theorem. This is a classic example of a combinatorics problem where we need to distribute indistinguishable objects into distinguishable bins with certain restrictions.

First, assign 2 apples to each child to meet the condition that each child gets at least 2 apples. Therefore, we distribute:

• Apples given initially to each child: apples

Now, we have apples left to distribute among the 3 children with no further restrictions.

According to the stars and bars method, the problem now is equivalent to finding the number of non-negative integer solutions to the equation:

where , , and are the number of additional apples given to the first, second, and third child, respectively.

The number of solutions is given by the formula for combinations with repetition, which is:

In our case, (apples left) and (children), so:

Calculate as follows:

Thus, the number of ways to distribute the apples under the given conditions is 136.

Therefore, the correct answer is 136.