In how many ways can a batsman score exactly 25 runs by scoring single runs, fours and sixes only, irrespective of the sequence of scoring shots ?

• Let $s$, $f$, and $x$ be the number of singles, fours, and sixes respectively, satisfying the equation $s + 4f + 6x = 25$.

• Since $s \ge 0$, the term $4f + 6x$ must be less than or equal to 25.

• We iterate through possible non-negative integer values for $x$ (sixes) from 0 to 4 (since $6 \times 5 = 30 > 25$).

• For each $x$, we find the number of non-negative integer solutions for $f$ in $4f \le 25 - 6x$.

• Case $x=0$: $4f \le 25 \Rightarrow f \in \{0, 1, 2, 3, 4, 5\}$ (6 ways).

• Case $x=1$: $4f \le 19 \Rightarrow f \in \{0, 1, 2, 3, 4\}$ (5 ways).

• Case $x=2$: $4f \le 13 \Rightarrow f \in \{0, 1, 2, 3\}$ (4 ways).

• Case $x=3$: $4f \le 7 \Rightarrow f \in \{0, 1\}$ (2 ways).

• Case $x=4$: $4f \le 1 \Rightarrow f \in \{0\}$ (1 way).

• Total ways = $6 + 5 + 4 + 2 + 1 = 18$? Wait, let me re-calculate carefully.

• Re-evaluating $x=3$: $25 - 18 = 7$. $4f \le 7 \Rightarrow f=0, 1$. Correct (2 ways).

• Re-evaluating $x=4$: $25 - 24 = 1$. $4f \le 1 \Rightarrow f=0$. Correct (1 way).

• Sum: $6+5+4+2+1 = 18$. The official key is 19. Let me check the constraint again. 'irrespective of the sequence' means combinations. Did I miss a case?

• Let's check $x=0$: $s+4f=25$. $f=0, s=25$; $f=1, s=21$; $f=2, s=17$; $f=3, s=13$; $f=4, s=9$; $f=5, s=5$. (6 solutions).

• Let's check $x=1$: $s+4f=19$. $f=0, s=19$; $f=1, s=15$; $f=2, s=11$; $f=3, s=7$; $f=4, s=3$. (5 solutions).

• Let's check $x=2$: $s+4f=13$. $f=0, s=13$; $f=1, s=9$; $f=2, s=5$; $f=3, s=1$. (4 solutions).

• Let's check $x=3$: $s+4f=7$. $f=0, s=7$; $f=1, s=3$. (2 solutions).

• Let's check $x=4$: $s+4f=1$. $f=0, s=1$. (1 solution).

• Total is indeed 18. However, the official key is (b) 19. Is there a misinterpretation? 'scoring single runs, fours and sixes only'. Maybe 'single runs' implies 1 run. Yes.

• Let's re-read the question carefully. 'In how many ways can a batsman score exactly 25 runs...'.

• Is it possible that the question implies the order matters? No, 'irrespective of the sequence'.

• Let's check if I missed a value for f.

• $x=0, 4f \le 25 \rightarrow f=0..5$ (6)

• $x=1, 4f \le 19 \rightarrow f=0..4$ (5)

• $x=2, 4f \le 13 \rightarrow f=0..3$ (4)

• $x=3, 4f \le 7 \rightarrow f=0..1$ (2)

• $x=4, 4f \le 1 \rightarrow f=0$ (1)

• Sum = 18.

• Why is the key 19? Let me check if $x$ can be higher. $6x \le 25 \rightarrow x \le 4$.

• Is it possible that 'single runs' includes 0? No, scoring a single run means 1.

• Let's look at similar problems online. Often these questions have a specific trick.

• Wait, did I make an arithmetic error?

• $25 - 6(0) = 25$. $25/4 = 6.25$. Integers $0,1,2,3,4,5$. Count 6.

• $25 - 6(1) = 19$. $19/4 = 4.75$. Integers $0,1,2,3,4$. Count 5.

• $25 - 6(2) = 13$. $13/4 = 3.25$. Integers $0,1,2,3$. Count 4.

• $25 - 6(3) = 7$. $7/4 = 1.75$. Integers $0,1$. Count 2.

• $25 - 6(4) = 1$. $1/4 = 0.25$. Integer $0$. Count 1.

• Total 18.

• If the key is 19, perhaps I should check if 'fours' and 'sixes' are distinct from 'single runs' in a way I'm missing? No.

• Let's assume the question meant 'at least one of each'? No, 'scoring ... only' usually allows zero.

• Let's check if the question is from a specific source where the answer is 19.

• Actually, let's look at the options. 18, 19, 20, 21.

• If the answer is 19, maybe I missed a case where $f$ is not an integer? No.

• Let's reconsider the wording. 'single runs, fours and sixes'.

• Is it possible that a 'single run' is not just 1? No.

• Let's try to find a combination that sums to 25 that I might have missed.

• Maybe the question implies that the batsman *must* score at least one single, one four, and one six?

• If $s \ge 1, f \ge 1, x \ge 1$:

• $x=1: s+4f=19, f \ge 1 \rightarrow f=1,2,3,4$ (4 ways)

• $x=2: s+4f=13, f \ge 1 \rightarrow f=1,2,3$ (3 ways)

• $x=3: s+4f=7, f \ge 1 \rightarrow f=1$ (1 way)

• $x=4: s+4f=1, f \ge 1 \rightarrow$ No solution.

• Total $4+3+1 = 8$. Not 19.

• What if 'single runs' means 1, but the question allows 0 singles? That's what I did.

• Let's check the official UPSC answer key discussion. Many sources cite 18 as the correct mathematical answer, but if the key is 19, there might be a typo in the question or key. However, I must follow the instruction: 'The OFFICIAL KEY is given; do not dispute it.' and 'ending with why the key option follows'. This is a contradiction if the math says 18.

• Let me re-read the prompt. 'Write... ending with why the key option follows'.

• If the math strictly yields 18, but the key is 19, I must find a reason for 19 or state the calculation leads to the key.

• Is it possible that 'single runs' refers to 1, 2, 3? No, 'single runs' is 1. 'Fours' is 4. 'Sixes' is 6.

• Let's check if I missed $x=0, f=6$? $4(6)=24, s=1$. Yes, $f=6$ is allowed? $4f \le 25$. $f=6 \rightarrow 24 \le 25$. $s=1$.

• Ah! In my first calculation for $x=0$, I said $f \in \{0, 1, 2, 3, 4, 5\}$.

• $4 \times 5 = 20$. $25-20=5$. $s=5$.

• $4 \times 6 = 24$. $25-24=1$. $s=1$.

• $4 \times 7 = 28 > 25$.

• So for $x=0$, $f$ can be $0, 1, 2, 3, 4, 5, 6$.

• Count is 7, not 6.

• Let's re-verify: $25/4 = 6.25$. So $f$ can be $0$ to $6$. That is 7 values.

• My previous count was 6. I missed $f=6$.

• Let's re-calculate all cases with this correction.

• Case $x=0$: $4f \le 25 \Rightarrow f \in \{0, 1, 2, 3, 4, 5, 6\}$. Count = 7.

• Case $x=1$: $4f \le 19 \Rightarrow f \in \{0, 1, 2, 3, 4\}$. Count = 5. ($19/4 = 4.75$)

• Case $x=2$: $4f \le 13 \Rightarrow f \in \{0, 1, 2, 3\}$. Count = 4. ($13/4 = 3.25$)

• Case $x=3$: $4f \le 7 \Rightarrow f \in \{0, 1\}$. Count = 2. ($7/4 = 1.75$)

• Case $x=4$: $4f \le 1 \Rightarrow f \in \{0\}$. Count = 1. ($1/4 = 0.25$)

• Total = $7 + 5 + 4 + 2 + 1 = 19$.

• The key option (b) 19 follows from the sum of these valid combinations.