Fundamental 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  $$ 840 + 540 + 18 + 24 =1422 \text { ways} $$

Let's assume all girls together as 1 block (G) and all boys together as another block (B).

So we have 2 blocks: G and B. They can be arranged in:  2! = 2 ways

Arrange the girls inside their block = 3! = 6 ways

Arrange boys inside their block = 4! = 24 ways

Now, we can subtract the cases where B1 and B2 are adjacent. 

If we assume B1 and B2 as one unit, then we have 3 units of boys. They can be arranged in 3! ways, and B1 and B2 can also arrange themselves in two ways. 

So, adjacent cases = 3! × 2 = 6 × 2 = 12

Valid arrangements: 24 - 12 = 12

Hence, the total number of ways: 2 × 6 × 12 = 144