How many such pairs of letters are there in t | vedclass

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(D) To find the number of such pairs,we compare the positions of letters in the word $EXCURSION$ with their positions in the English alphabet.Word: $E

Vedclass · Accepted Answer

More than three

Vedclass · Answer

(D) To find the number of such pairs,we compare the positions of letters in the word $EXCURSION$ with their positions in the English alphabet.Word: $E, X, C, U, R, S, I, O, N$Positions: $5, 24, 3, 21, 18, 19, 9, 15, 14$$1$. Checking forward:- $E$ to $X$: $5$ to $24$ (No)- $E$ to $C$: $5$ to $3$ (No)- $E$ to $U$: $5$ to $21$ (No)- $E$ to $R$: $5$ to $18$ (No)- $E$ to $S$: $5$ to $19$ (No)- $E$ to $I$: $5$ to $9$ (No)- $E$ to $O$: $5$ to $15$ (No)- $E$ to $N$: $5$ to $14$ (No)- $X$ to $C$: $24$ to $3$ (No)- $C$ to $U$: $3$ to $21$ (No)- $C$ to $R$: $3$ to $18$ (No)- $C$ to $S$: $3$ to $19$ (No)- $C$ to $I$: $3$ to $9$ (No)- $C$ to $O$: $3$ to $15$ (No)- $C$ to $N$: $3$ to $14$ (No)- $U$ to $R$: $21$ to $18$ (No)- $U$ to $S$: $21$ to $19$ (No)- $U$ to $I$: $21$ to $9$ (No)- $U$ to $O$: $21$ to $15$ (No)- $U$ to $N$: $21$ to $14$ (No)- $R$ to $S$: $18$ to $19$ (Yes,$1$ letter between them in alphabet is $0$)- $R$ to $I$: $18$ to $9$ (No)- $R$ to $O$: $18$ to $15$ (No)- $R$ to $N$: $18$ to $14$ (No)- $S$ to $I$: $19$ to $9$ (No)- $S$ to $O$: $19$ to $15$ (No)- $S$ to $N$: $19$ to $14$ (No)- $I$ to $O$: $9$ to $15$ (No)- $I$ to $N$: $9$ to $14$ (No)- $O$ to $N$: $15$ to $14$ (Yes,$1$ letter between them in alphabet is $0$)$2$. Checking backward:- $N$ to $O$: $14$ to $15$ (Yes,$1$ letter between them in alphabet is $0$)- $N$ to $I$: $14$ to $9$ (No)- $N$ to $S$: $14$ to $19$ (No)- $N$ to $R$: $14$ to $18$ (No)- $N$ to $U$: $14$ to $21$ (No)- $N$ to $C$: $14$ to $3$ (No)- $N$ to $X$: $14$ to $24$ (No)- $N$ to $E$: $14$ to $5$ (No)- $O$ to $I$: $15$ to $9$ (No)- $O$ to $S$: $15$ to $19$ (No)- $O$ to $R$: $15$ to $18$ (No)- $O$ to $U$: $15$ to $21$ (No)- $O$ to $C$: $15$ to $3$ (No)- $O$ to $X$: $15$ to $24$ (No)- $O$ to $E$: $15$ to $5$ (No)- $I$ to $S$: $9$ to $19$ (No)- $I$ to $R$: $9$ to $18$ (No)- $I$ to $U$: $9$ to $21$ (No)- $I$ to $C$: $9$ to $3$ (No)- $I$ to $X$: $9$ to $24$ (No)- $I$ to $E$: $9$ to $5$ (No)- $S$ to $R$: $19$ to $18$ (Yes,$1$ letter between them in alphabet is $0$)- $S$ to $U$: $19$ to $21$ (No)- $S$ to $C$: $19$ to $3$ (No)- $S$ to $X$: $19$ to $24$ (No)- $S$ to $E$: $19$ to $5$ (No)- $R$ to $U$: $18$ to $21$ (No)- $R$ to $C$: $18$ to $3$ (No)- $R$ to $X$: $18$ to $24$ (No)- $R$ to $E$: $18$ to $5$ (No)- $U$ to $C$: $21$ to $3$ (No)- $U$ to $X$: $21$ to $24$ (No)- $U$ to $E$: $21$ to $5$ (No)- $C$ to $X$: $3$ to $24$ (No)- $C$ to $E$: $3$ to $5$ (Yes,$1$ letter between them in alphabet is $1$)- $X$ to $E$: $24$ to $5$ (No)Total pairs: $(R, S), (O, N), (N, O), (S, R), (C, E)$. There are $5$ pairs. Since $5 > 3$,the answer is More than three.

How many such pairs of letters are there in the word $EXCURSION$,each of which has as many letters between them in the word as they have in the English alphabet?

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