Addition Principle of Counting

$$p$$ is the number of elements $$(m\ ,\ n)$$ such that $$6^m+9^n$$ is a multiple of 5.

$$6^m+9^n=0$$ (mod 5) 

6 = 1 (mod 5) 

$$6^m=1^m=1$$ (mod 5) 

9 = 4 (mod 5) 

$$9^n=4^n$$ (mod 5) 

$$1+4^n=0$$ (mod 5) 

Now, $$4^1=4$$ (mod 5)

$$4^2=1$$ (mod 5)

$$4^3=4$$ (mod 5)

$$4^4=1$$ (mod 5)

Now, we will get a remainder of 1 if we divide 1 by 5 (for any value of $$m$$) and a remainder of either 1 or 4 from $$4^n$$ if $$n$$ is even or odd, respectively. 

It is given that $$6^m+9^n$$ is a multiple of 5. 

It means that the remainder that we get from $$6^m+9^n$$ must be a multiple of 5. 

So, we need the remainder of 4 from $$4^n$$. 

Hence, $$n$$ will be odd. 

$$m$$ can take any value from 1 to 50 and $$n$$ can only take odd numbers. 

So, the number of elements of $$(m\ ,\ n)$$ for which $$6^m+9^n$$ must be a multiple of 5 is $$\left(50\times25\right)=1250$$

$$p$$ = 1250 

Now, $$q$$ is the number of elements $$(m\ ,\ n)$$ such that $$(m+n)$$ is a square of a prime number. 

$$1\le\left(m\ ,\ n\right)\le50$$

$$2\le\left(m\ +\ n\right)\le100$$

Square of prime numbers which is less than 100 = $$2^2,\ 3^2,\ 5^2,\ 7^2$$

So, $$(m+n)$$ = 4, 9, 25 and 49

For $$1\le\left(m\ ,\ n\right)\le50$$, if $$(m+n)\le51$$, there will be a total of $$(m+n-1)$$ elements.  

For $$(m+n)$$ = 4, $$(m\ ,\ n)$$ = (1,3) , (2,2) , (3,1) i.e. 3 combinations. 

For $$(m+n)$$ = 9, $$(m\ ,\ n)$$ will have 8 combinations.

For $$(m+n)$$ = 25, $$(m\ ,\ n)$$ will have 24 combinations.

For $$(m+n)$$ = 49, $$(m\ ,\ n)$$ will have 48 combinations.

Total combinations of $$(m+n)$$ = $$(3+8+24+48)$$ = 83

$$q$$ = 83

So, $$(p+q)$$ = $$1250+83$$ = 1333

Hence, the sum of the values of $$p$$ and $$q$$ is 1333. 

The word given is PQRPQRSTUVP. Here, we have P 3 times, Q 2 times, R 3 times, and S,T,U,V one time each
Total distinct letters: 7

We form 4-letter words using these.

Case 1: All letters are distinct

Here, 4 distinct letters can be chosen from 7 letters in  $$ {}^7C_4 = 7 \cdot 6 \cdot 5 \cdot 4 = 840 \text { ways} $$

Case 2: One pair + two distinct

To choose a pair, we need to pick one from the letters that occur 2 times or more in the word. Of the letters P,Q and R occure more than once, so the letter that will form the pair can be picked in 3 ways. For the remaining 2 distinct letters, they can be picked from the remaining 6 letters in  $$  \binom{6}{2} = 15 $$ ways. These can be arranged amongst themselves in  $$  \frac{4!}{2!} = 12$$ ways

Total number of ways words can be formed is= $$ 3 \cdot 15 \cdot 12 = 540 \text { ways} $$

Case 3: Two pairs

Here, we need to choose 2 letters (each must occur ≥2 times): from (P,Q,R), which can be done in 3 ways. The letters can be arranged amongst themselves in $$ \dfrac{4!}{2!2!} = 6 $$ ways

Total number of ways words can be formed is $$ 3 \cdot 6 = 18 \text{ ways} $$

Case 4: Three same + one different

Only P occurs three times, so here it can be picked in 1 way, and the 1 different letter from the remaining 6 can be picked in 6 ways. The letters can be arranged amongst themselves in  $$ \dfrac{4!}{3!} = 4  $$

Total number of ways words can be formed is $$ 1 \cdot 6 \cdot 4 = 24 \text{ ways}$$

Hence, the total number of ways words can be formed considering all the cases is  $$ 840 + 540 + 18 + 24 =1422 \text { ways} $$

The digits given are $${0,1,2,5,9} $$.
We need to make 4-digit numbers that are $$>5000$$ and $$<9000$$, and are divisible by 3.

Thus, the number is of the form $$5abc$$. The first numbe rhas to be 5 otherwise the numbers formed will not be in the given range.

For divisibility by 3 $$ 5+a+b+c \equiv 0 \pmod{3} $$ , essentially sum of the digits should be either 0 or a multiple of 3. 

We form the following cases: 

Case 1: All digits different

Possible set of selection:  $$ (5,0,1,9), \quad \text{sum } = 15 \equiv 0 $$

The place of 5 is fixed, the remaining three digits can be arranged in $$ 3! = 6 $$

Hence, 6 possible numbers can be obtained. 

Case 2: Two alike, two different

If 5 is the number that is repeated, then the possible selections are: $$ (5,5,0,2),\ (5,5,2,9) $$

Since the 5 at the first place is fixed, in each of these, the numbers can be arranged in $$ 3! = 6 \text{ ways} $$ 

Otherwise, if 5 is not the number repeated then the valid selections are $$(5$$, $$0$$, $$0$$, $$1)$$, $$\ (5$$, $$1$$, $$1$$, $$2)$$, $$\ (5$$, $$2$$, $$2$$, $$0)$$, $$\ (5$$, $$2$$, $$2$$, $$9)$$, $$\ (5$$, $$1$$, $$9$$, $$9)$$

In these cases, the digits can be arranged in $$ \frac{3!}{2!} = 3 \text{ ways} $$

So, the total number of numbers obtained in this case is $$ 5 \cdot 3 + 2 \cdot 6 = 15 + 12 = 27 $$

Case 3: Three alike, one different

If 5 is the digit that is repeated 3 times, the valid selections are: $$ (5,5,5,0),\ (5,5,5,9) $$

The digits can be arranged in $$ \frac{3!}{2!} = 3 \text{ ways} $$

If any other digit other than 5 is repeated 3 times, then we can have numbers 5000, 5222 and 5999 none of which are divisible by 3. hence they are not valid choices. 

So, the total number of numbers obtained in this case is $$3+3=6 \text{ ways}$$

Case 4: Two pairs

The only valid arrangement here is $$ (5,5,1,1) $$ , and the digits can be arranged in  $$ \frac{3!}{2!} = 3 $$. So the total number of numbers obtained is 3. 

Hence, the total number of such numbers is  $$ 6 + 27 + 6 + 3 = 42 $$