Let's take the number of paths between Q and R to be b and the number of paths between R and S to be a

We are given the paths from P to S through R (which would be 4a), the paths from P to S through Q (which would be 12) and the paths from P to Q to R to S, which would be 3ab) is equal to 62

Giving the relation 4a+12+3ab = 62

Or 4a+3ab = 50

The paths from Q to R directly (which would be b), through P( which would be 12) and through S (which would be 4a) are 27

Giving the relation b+12+4a = 27

Or 4a+b = 15

Subtracting this equation from the first one we got, we get 3ab-b=35, or b(3a-1)=35

b can be 1, 3, 5 or 7

Substituting these values in the second equation, we see that it can not be 1 or 5, leaving only 3 or 7 as the possible values. 

Substituting b as 3 in the first equation would give 13a=50, which is not true. 

Substituting b as 7 in the first equation would give 25a= 50, which would give a=2

We are asked the number of paths from Q to R, which is b=7

Therefore, 7 is the correct answer. 

Let the number of balloons each child received be 2a, 2b, 2c and 2d

2a + 2b + 2c + 2d = 20

a + b + c + d = 10

Each of them should get more than zero balloons.

Therefore, total number of ways = $ (n-1)C_{r-1} = (10-1)C_{4-1} = 9C_3 = 84 $

Case 1: 4-digit numbers

Given digits - 0, 1, 2, 3, 4, 5

_, _, _, _

As the numbers should be greater than 2000, first digit can be 2, 3, 4 and 5.

For remaining digits, we need to arrange 3 digits from the remaining 5 digits, i.e. 5*4*3 = 60 ways

Total number of possible 4-digit numbers = 4*60 = 240

Case 2: 5-digit numbers

_, _, _, _, _

First digit cannot be zero.

Therefore, total number of cases = 5*5*4*3*2 = 600

Case 3: 6-digit numbers

_, _, _, _, _, _

First digit cannot be zero.

Therefore, total number of cases = 5*5*4*3*2*1 = 600

Total number of integers possible = 600 + 600 + 240 = 1440

The answer is option A.

The number of 4-digit numbers possible using 1,1,2, and 4 is

$\frac{4!}{2!} = 12$

Number of 1's, 2's and 4's in units digits will be in the ratio 2:1:1, i.e. 6 1's, 3 2's and 3 4's.

Sum = $6(1) + 3(2) + 3(4) = 24$

Similarly, in tens digit, hundreds digit and thousands digit as well.

Therefore, sum = $24 + 24(10) + 24(100) + 24(1000) = 24(1111)$

Mean = $\frac{24(1111)}{12} = 2222$

The possible arrangements are of the form 

35 _ Can be chosen in 6 ways.

35 _ _ We can choose 2 out of the remaining 6 in 6C2=15 ways.

We remove 1 case where 7 and 8 are together to get 14 ways.

35 _ _ _We can choose 3 out of the remaining 6 in 6C3=20 ways.

We remove 4 cases where 7 and 8 are together to get 16 ways.

35 _ _ _ _We can choose 4 out of the remaining 6 in 6C4=15ways.

We remove 6 case where 7 and 8 are together to get 9 ways.

35 _ _ _ _ _ We choose 1 out of 7 and 8 and all the remaining others in 2 ways.

Thus, total number of cases = 6+14+16+9+2 = 47.

Alternatively,

The arrangement requires a selection of 3 or more numbers while including 3 and 5 and 7, 8 are never included together. We have cases including a selection of only 7, only 8 and neither 7 nor 8.

Considering the cases, only 7 is selected.

We can select a maximum of 7 digit numbers. We must select 3, 5, and 7.

Hence we must have ( 3, 5, 7) for the remaining 4 numbers we have

Each of the numbers can either be selected or not selected and we have 4 numbers :

Hence we have _ _ _ _ and 2 possibilities for each and hence a total of 2*2*2*2 = 16 possibilities.

SImilarly, including only 8, we have 16 more possibilities.

Cases including neither 7 nor 8.

We must have 3 and 5 in the group but there must be no 7 and 8 in the group.

Hence we have 3 5 _ _ _ _.

For the 4 blanks, we can have 2 possibilities for either placing a number or not among 1, 2, 4, 6.

= 16 possibilities

But we must remove the case where neither of the 4 numbers are placed because the number becomes a two-digit number.

Hence 16 - 1 = 15 cases.

Total = 16+15+16 = 47 possibilities

Let the numbers be of the form 100a+10b+c, where a, b, and c represent single digits.

Then (100c+10b+a)-(100a+10b+c)=198

99c-99a=198

c-a = 2.

Now, a can take the values 1-7. a cannot be zero as the initial number has 3 digits and cannot be 8 or 9 as then c would not be a single-digit number.

Thus, there can be 7 cases.

B can take the value of any digit from 0-9, as it does not affect the answer. Hence, the total cases will be 7× 10=70

Event 1: Distribution of balloons

Since each child gets at least 4 balloons, we will initially allocate these 4 balloons to each of them.

So we are left with $15 - 4 \times 3 = 15 - 12 = 3$ balloons and 3 children.

Now we need to distribute 3 identical balloons to 3 children.

This can be done in ${}^{n+r-1}C_{r-1}$ ways, where $n = 3$ and $r = 3$.

So, number of ways = ${}^{3+3-1}C_{3-1} = {}^5C_2 = \frac{5 \times 4}{2 \times 1} = 10$

Event 2: Distribution of pencils

Since each child gets at least one pencil, we will allocate 1 pencil to each child. We are now left with $6 - 3 = 3$ pencils.

We now need to distribute 3 identical pencils to 3 children.

This can be done in ${}^{n+r-1}C_{r-1}$ ways, where $n = 3$ and $r = 3$.

So, number of ways = ${}^{3+3-1}C_{3-1} = {}^5C_2 = \frac{5 \times 4}{2 \times 1} = 10$

Event 3: Distribution of erasers

We need to distribute 3 identical erasers to 3 children.

This can be done in ${}^{n+r-1}C_{r-1}$ ways, where $n = 3$ and $r = 3$.

So, number of ways = ${}^{3+3-1}C_{3-1} = {}^5C_2 = \frac{5 \times 4}{2 \times 1} = 10$

Applying the product rule, we get the total number of ways = $10 \times 10 \times 10 = 1000$.

The question asks for the number of 4 digit numbers using only the digits 1, 2, and 3 such that the digits 2 and 3 appear at least once.

The different possibilities include :

Case 1:The four digits are ( 2, 2, 2, 3). Since the number 2 is repeated 3 times. The total number of arrangements are :

$\frac{4!}{3!​}$ = $4$.

Case 2: The four digits are 2, 2, 3, 3. The total number of four-digit numbers formed using this are :

$\frac{4!}{2!(2!)​}$ = $6$.

Case 3: The four digits are 2, 3, 3, 3. The number of possible 4 digit numbers are :

$\frac{4!}{3!​}$ = $4$.

Case4: The four digits are 2, 3, 3, 1. The number of possible 4 digit numbers are :

$\frac{4!}{2!​}$ = $12$.

Case5: Using the digits 2, 2, 3, 1. The number of possible 4 digit numbers are :

$\frac{4!}{2!​}$ = $12$.

Case 6: Using the digits 2, 3, 1, 1. The number of possible 4 digit numbers are :

$\frac{4!}{2!​}$ = $12$.

A total of 12 + 12 + 12 + 4 + 6 + 4 = 50 possibilities.

Alternatively,

We have to form 4 digit numbers using 1,2,3 such that 2,3 appears at least once 

So the possible cases :

Now we get $\frac{4!}{2!​}$ × 3 = 36 ( When one digit is used twice and the remaining two once )

$\frac{4!}{3!​}$ × 2 = 8 ( When 1 is used 0 times and 2 and 3 is used 3 times or 1 time )

$\frac{4!}{2!​ × 2!}$ = 6( When 2 and 3 is used 2 times each )

So total numbers = 36+8+6 = 50

The number of paths from (1, 1) to (8, 10) via (4, 6) = The number of paths from (1,1) to (4,6) * The number of paths from (4,6) to (8,10)

To calculate the number of paths from (1,1) to (4,6), 4-1 =3 steps in x-directions and 6-1=5 steps in y direction

Hence the number of paths from (1,1) to (4,6) = ${}^{(3+5)}C_3​$ = 56

To calculate the number of paths from (4,6) to (8,10), 8-4 =4 steps in x-directions and 10-6=4 steps in y direction

Hence the number of paths from (4,6) to (8,10) = ${}^{(4+4)}C_4$​ = 70

The number of paths from (1, 1) to (8, 10) via (4, 6) = 56*70=3920