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(B) To find the number of pairs of letters in the word $REPURCUSSION$ that have the same number of letters between them as in the English alphabet,we

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(B) To find the number of pairs of letters in the word $REPURCUSSION$ that have the same number of letters between them as in the English alphabet,we check the sequence of letters:$1$. $R$ $(18)$ to $U$ $(21)$: There are two letters $(E, P)$ between them in the word. In the alphabet,$S, T$ are between $R$ and $U$. (No)$2$. $P$ $(16)$ to $S$ $(19)$: There are two letters $(U, R)$ between them in the word. In the alphabet,$Q, R$ are between $P$ and $S$. (No)$3$. $R$ $(18)$ to $U$ $(21)$ (second $R$): There are two letters $(C, U)$ between them in the word. In the alphabet,$S, T$ are between $R$ and $U$. (No)$4$. $I$ $(9)$ to $N$ $(14)$: There are four letters $(O, N)$ - wait,let's check forward: $R(18), E(5), P(16), U(21), R(18), C(3), U(21), S(19), S(19), I(9), O(15), N(14)$.Checking pairs:- $P(16)$ and $S(19)$: $P, U, R, C, U, S$. Between $P$ and $S$ are $U, R, C, U$ ($4$ letters). In alphabet,$Q, R$ ($2$ letters). (No)- $I(9)$ and $N(14)$: Between $I$ and $N$ is $O$ ($1$ letter). In alphabet,$J, K, L, M$ ($4$ letters). (No)- $R(18)$ and $U(21)$: $R, E, P, U$. Between $R$ and $U$ are $E, P$ ($2$ letters). In alphabet,$S, T$ ($2$ letters). (Yes,one pair: $R-U$)- $O(15)$ and $N(14)$: Adjacent. (No)- $S(19)$ and $U(21)$: $S, S, I, O, N$ (No).After careful analysis,there is only one such pair: $(R, U)$.

How many pairs of letters are there in the word $REPURCUSSION$ which have as many letters between them in the word as in the alphabet and that too in the same order?

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