Number, Letter and Symbol Arrangement

What the exam asks. A long series of single-digit numbers (or of letters) is printed, and the question reads: 'How many such even digits are there, each of which is immediately preceded by an odd digit and also immediately followed by an odd digit?' Counting is from left to right only. This is the most repeated form in the whole chapter; such questions are asked regularly in Railway papers.

The logic. The element you count has to pass two tests at once, one on each side. So it needs a left neighbour and a right neighbour, and that is why the first and the last cell of the strip can never be counted. One matching side is not enough.

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Step method.
1. Write a code line under the strip: O or E for digits, V or C for letters.
2. Turn the condition into a three-letter pattern in the order the question gives. 'Even, preceded by odd, followed by odd' becomes O E O. The middle letter is the element you count.
3. Find the pattern in the code line and tick its middle cell.
4. Count the ticks.

In the picture the strip is 2 5 8 3 6 4 7 9 4 1 and its code line is E O E O E E O O E O. The pattern O E O appears around the 8 (place 3) and around the 4 (place 9). The 6 fails on the right (a 4 follows it), the first 4 fails on the left (a 6 precedes it), and the 2 at place 1 cannot be counted at all. The answer is 2.

Matches may overlap. In O E O E O both E cells are counted, because each one is the middle of its own pattern. The counted element is the middle of the pattern, and the two ends of the strip never qualify.

Letters. For vowel questions write V for A E I O U and C for every other letter. Remember that Y is a consonant.

What the exam asks. The strip now mixes digits, letters and symbols, for example # $ Ω 2 1 @ 4 & ~ % 8 6 9 + 5 ? £ 3 7 * =, and the condition names kinds: 'How many such symbols are there, each of which is immediately preceded by an odd number but not immediately followed by an even number?' This is the standard question of the Graduate-level Railway papers.

The logic. The condition does not care about the exact digit or the exact symbol, only about its kind (and sometimes one property of that kind). So translate the strip into kind-codes: N number, L letter, S symbol. If a property also matters, use O / E for numbers and V / C for letters.

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In the picture the strip 6 @ 2 9 # K 7 $ 4 R 8 % 1 T has the code line N S N N S L N S N L N S N L. The question 'how many numbers are immediately followed by a symbol?' is just the pair N S. The red numbers 6, 9, 7 and 8 are followed by symbols (blue); the 2, the 4 and the 1 are followed by a digit, a letter and a letter, so they fail.

Method.
1. Write the kind-code line under the strip.
2. Turn the condition into a box of codes in the order given, such as [N][S] or [S][N][S].
3. Slide the box on the code line and tick the element the question names.
4. Count the ticks.

Patterns of three. Some questions stack two conditions on one side, or put one on each side: 'a symbol preceded by two numbers', 'a letter preceded by a symbol and followed by a number'. Build the three-cell box from kinds and slide it.

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In the picture the symbols that follow two numbers are #, @, % and & (green brackets). The $ fails because only one number stands before it.

'But not'. 'Not followed by an even number' means followed by an odd number, a letter or a symbol. Count the larger group first and remove the unwanted part (see the Kavach Trick).

Code check. A short variant gives four codes and asks which one breaks the pattern, for example 'two digits followed by two letters' in 11AB, 22CD, E34F, 55GH. Write N / L under each: NNLL, NNLL, LNNL, NNLL. The odd one out is E34F.

What the exam asks. Four small forms live here, all on a series of single digits or letters.
1. A direct count: 'How many vowels are there in the series?'
2. A unique count: 'How many unique digits are odd and greater than 5 (each counted only once, regardless of repetitions)?'
3. A one-sided neighbour count: 'How many letters are immediately followed by a vowel?'
4. A position: 'What is the position of the letter L from the left end?'

The logic. These are the simple cousins of the both-side count. The box has only two cells, [X][Y], and you count the cell the question names. For a direct count you simply tick the cells that have the property.

Unique count. The words 'counted only once' change the job. Repeats no longer matter, so do not walk through the strip counting. Instead list the candidates, the digits that have the property, and ask for each one: 'is it present in the strip?'

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In the strip 7 2 4 5 9 1 6 3 8 7 4 2 6 1 5 9 3 7 4 8 2 1, the digits that are odd and greater than 5 are 7 and 9. Both are present (red), so the answer is 2. Note that 5 is not greater than 5, and 0 is absent.

One-sided neighbour count. For 'followed by' the box is [counted][next]; for 'preceded by' it is [previous][counted]. In a series made only of letters, 'letters immediately followed by a vowel' is simply the number of vowels that do not stand at place 1, because each such vowel has exactly one letter before it.

Position. The n-th element from the left is place n. The n-th from the right is place (L − n + 1) from the left, where L is the length of the strip. Unique means list the candidates; one-sided means count the cell that is named.

What the exam asks. Three forms, all on the same idea.
1. 'Each digit of 6534217 is arranged in ascending order from left to right. In how many places does the digit remain unchanged?'
2. 'What is the sum of the digits that are third from the left and second from the right in the new number?'
3. Five three-digit numbers are given: 'Add the third digit of the highest number to the second digit of the lowest number', or 'If the numbers are arranged in ascending order, the position of how many remains unchanged?'

The logic. Ascending order puts the smallest on the left. So a digit belongs at the place equal to its rank: 1 plus the number of smaller digits. It stays where it is only when its rank equals the place it already occupies. If a digit repeats, let s digits be smaller and c digits be equal to it (itself included): it can sit at any place from s + 1 to s + c, and it stays if its present place is in that range (in 122 both 2s stay). The word that matters is place: looking 'in order' with the neighbours is not enough.

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The top row is 6 2 3 8 5 1 7 4 and the bottom row is the ascending order 1 2 3 4 5 6 7 8. Compare column by column. Places 2, 3, 5 and 7 are equal (green), so four digits keep their places. The 5 at place 5 has exactly four smaller digits (1, 2, 3, 4), which is why it stays.

Method. Write the sorted row under the number, tick the equal columns, count the ticks. For descending order, count the larger digits instead of the smaller ones.

Picking a place without sorting everything. In the new number the third digit from the left is the 3rd smallest digit, and the second digit from the right is the 2nd largest. Pick them out directly.

Three-digit numbers. The highest number has the largest hundreds digit; if two tie, the tens digit decides, then the units digit. The lowest is the opposite. For 538, 291, 764, 413, 685 the highest is 764 and the lowest is 291, so the third digit of the highest (4) plus the second digit of the lowest (9) is 13. 'Unchanged position' for numbers uses the same rank test on whole numbers.

What the exam asks. 'If 2 is added to each odd digit and 2 is subtracted from each even digit of the number 7213546, what is the difference between the largest and the smallest digit of the new number?' An operation changes every digit, and the question is about the new number: highest minus lowest, a digit at a given place from an end, a sum, 'how many digits appear more than once', or 'the sum of the repeated digits'. A letter form also comes: 'each vowel is changed to the next letter and each consonant to the previous letter; how many clusters have no vowel?'

The logic. The operation depends on the value of a digit (odd or even), never by its place. Once the new number is written, the old one is finished with. Every error in this type comes from mixing the two.

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In the picture the number is 3 6 8 1 4 7 and the operation is 'odd digit + 1, even digit − 2'. The odd digits 3, 1, 7 (blue) become 4, 2, 8; the even digits 6, 8, 4 (orange) become 4, 6, 2. The new number is 4 4 6 2 2 8.

Method.
1. Write the new number directly below the old one, digit under digit.
2. Answer from the new number only.
3. For 'appear more than once', tally the new digits. In 4 4 6 2 2 8 the digits 4 and 2 repeat, so the answer is 2 digits (not 4 places). For the sum of repeated digits add each repeating digit once: 4 + 2 = 6.

Which digits become odd or even? Odd ± odd and even ± even give even; odd ± even and even ± odd give odd. So with 'odd digit + 1 and even digit − 2', every new digit is even.

Letters and tables. The same method works on letters: vowels and consonants get different shifts, and you write the new word under the old one. If a table says 1 = R, 2 = A, …, convert the whole digit string into letters first, then answer on the letters.

What the exam asks. A mixed series of numbers, letters and symbols is printed and the question says: 'If all the symbols are dropped from the series, which of the following will be fifth from the left?' or 'If all the letters are dropped, which will be seventh from the right?' The options are single elements.

The logic. After the removal the strip closes up and becomes shorter. Every element that is left has a new place number, so the place numbers of the old strip are of no use. All you need is the surviving elements in their old order.

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The strip is T 5 $ A 8 # 3 K @ 6 E % (12 elements). Remove the symbols $, #, @ and %. The new strip is T 5 A 8 3 K 6 E, with 8 elements. From the left the third element is A. From the right the elements are E, 6, K, 3, 8, A, 5, T, so the fifth from the right is 8.

Method.
1. Copy only the survivors, in order. Do not cross out and count in the old strip.
2. Count the new length L.
3. Count from the end the question names. If you must change the end, the n-th from the right is the (L − n + 1)-th from the left.

Two kinds removed. If the question drops two kinds ('all vowels and all symbols'), remove both in the same pass. If it drops a single digit, remove every copy of it.

What the exam asks. Three forms share one idea.
1. A swap chain: 'The 1st and 4th digits are interchanged, then the 2nd and 5th, then the 3rd and 8th. Which digit is fourth from the right now?'
2. Machine steps: 'Input: 45, 12, 78, 36, 59. Step 1: 12 goes to the first place and 78 goes to the last place. Step 2: 36 goes to the second place and 59 goes to the fourth place. What is the final output?'
3. A word shuffle: 'LONG - ONLG - GNOL; EARN - AREN - NRAE. Which option follows the same pattern?'

The logic. Every step works on the strip as it stands at that moment, never on the original. That one idea decides all three forms.

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Start with 6 3 9 1 5 8 2 4. Swap places 1 and 4: 1 3 9 6 5 8 2 4. Now swap places 2 and 5 of this strip: 1 5 9 6 3 8 2 4. Now swap places 3 and 8: 1 5 4 6 3 8 2 9. The red cells are the pair that moved in that step. In the final strip the digits from the right are 9, 2, 8, 3, 6, 4, 5, 1, so the fourth from the right is 3.

Machine steps. Read each step as a change of place. Ask what the steps are building: here step 1 puts the smallest number (12) first and the largest (78) last, and step 2 puts the next smallest (36) second and the next largest (59) fourth. The machine is sorting the row, so the output is 12, 36, 45, 59, 78.

Word shuffle. Number the places of a word 1 2 3 4 and write what each step does as a place map. LONG → ONLG takes places 2, 3, 1, 4. ONLG → GNOL takes places 4, 2, 1, 3. The map must fit both given pairs; then apply it to the options.

Method for a swap chain. Rewrite the whole strip after every swap and mark the pair that moved. Answer from the last strip only.

What the exam asks. The 2026 Railway papers brought a new question: 'How many letters of the English alphabet lie between the 5th letter from the right end and the 4th letter from the left end of the series?' A longer-known form asks for pairs with a gap: 'how many 5s have a 3 exactly two places to their right (exactly one element between them)?'

The logic of the new pattern. The series is used only to pick two letters. Once they are picked, forget the series. The count is made on the alphabet A to Z, with A = 1 and Z = 26. The letters strictly between two letters of rank r₁ and r₂ number |r₁ − r₂| − 1. If the answer is 0, the letters are neighbours.

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Take the series Z K J H G F D S T U I A W Q P L M N B X. Counting from the right, the 5th letter is L (X, B, N, M, L). Counting from the left, the 4th letter is H (Z, K, J, H). In the alphabet H = 8 and L = 12, so 12 − 8 − 1 = 3 letters (I, J, K) lie between them.

Method.
1. Count from the right end to the first letter, and from the left end to the second. You never need the length of the series.
2. Write the two alphabet ranks.
3. Subtract and subtract 1.

Fixed gap (the older form). When the question says 'exactly k elements between', the places differ by k + 1.

[[fig:1]]

In 4 9 6 4 2 9 4 7 9 1, the 4s are at places 1, 4, 7. A 9 stands two places to the right of the 4 at place 4 (place 6) and of the 4 at place 7 (place 9), but not of the 4 at place 1 (place 3 holds 6). The answer is 2. Pick the letters in the series, then count in the alphabet; exactly k between means the places differ by k + 1.

What the exam asks. A 5 × 5 table of letters is printed with its rows and columns numbered 0 to 4, and the questions are: 'Which letters appear in (0,1), (3,0) and (1,2)?', 'Based on the matrix, decode (3,0)(2,3)(4,0)(0,4)', or 'What is the position of the letter K?'. An older picture style asks for 'the letter in the row of S and the column of C'.

The logic. A cell is named (row, column). The first number is the row, counted down from the top; the second is the column, counted across from the left. The numbers printed on the grid are the truth. In this paper they begin at 0, so (0,0) is the top-left cell, not (1,1).

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In the picture the letter K stands at (2,2), (3,1) and (4,0): one letter, three cells. This is why a 'position of K' question needs a full scan.

Method.
1. Read where the numbering starts (0 or 1).
2. For each pair, go down to the row first, then across to the column.
3. For 'position of X', scan row by row and list every cell holding X, because letters repeat.
4. For a decode, write the letters in the order in which the pairs are given.

Row of / column of. For 'the letter in the row of S and the column of C', take the row number from S and the column number from C, then read the crossing cell. The cell of S itself and the cell of C itself are not the answer.

What the exam asks. Five words are numbered 1 to 5, for example 1 Measurement, 2 Methodology, 3 Merchandise, 4 Meaningless, 5 Memorabilia, and the question asks for the order in which they appear in a dictionary, written as a sequence of numbers (here 4, 1, 5, 3, 2). The same skill is taught in the chapter on Alphabet, Dictionary Order and Word Formation.

The logic. A dictionary compares two words letter by letter from the left. The first place where they differ decides which word comes first; the letters before it are equal and tell you nothing. If one word is the beginning of the other, the shorter word comes first (Mimic before Mimicry).

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Measurement and Meaningless share M E A. The next letters are S and N. N comes before S, so Meaningless (4) comes before Measurement (1).

Method.
1. Cross out the beginning that all the words share (here 'Me').
2. Write the next letter of each word beside it and group the words: A (Measurement, Meaningless), M (Memorabilia), R (Merchandise), T (Methodology).
3. Order the groups by that letter. Inside a group go on to the next letter: Mea-ningless before Mea-surement.

The result is 4, 1, 5, 3, 2.

Letters of one word. A rarer form arranges the letters of a single word in alphabetical order and asks how many keep their place. RADHIKA becomes A A D H I K R.

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Compare place by place: R / A differ, A / A, D / D, H / H, I / I, K / K agree, A / R differ. Five places agree. Carry the repeated letter into the sorted row, or every later place slips.

What the exam asks. Two shapes of letter series sit here.
1. A cluster series: 'HLG, KOJ, NRM, QUP, ?' or a pair series such as 'JK - LM - NO'.
2. A repeating-block series: 'Sr_ss_s_s_ss' with four blanks, filled from the options in order, or a question like 'which letter is in the 23rd place of the repeating block?'. Both are also met in the chapter on Series.

Cluster logic. Each place inside the cluster is its own series. Turn the letters into numbers (A = 1 … Z = 26) and read the first letters, the second letters and the third letters separately.

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H K N Q move by +3, L O R U by +3 and G J M P by +3, so the next cluster is T X S. Some places follow other families: in UME, OPG, ISJ, EVN the first letters U, O, I, E are the vowels in reverse order, so the next is A; the second letters M, P, S, V move by +3, so the next is Y; the third letters E, G, J, N have gaps 2, 3, 4, so the next gap is 5 and the letter is S. The answer is AYS.

Block logic. A repeating series is a short block written again and again. If the block has L letters, the letter at place n depends only on the remainder when n is divided by L. A remainder of 0 means the last letter of the block.

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In P Q R P S P Q R P S … the block has L = 5 letters. The 17th letter: 17 ÷ 5 leaves 2, so it is the 2nd letter of the block, Q. The 20th letter leaves 0, so it is the last letter, S.

Blanks. In 'Sr_ss_s_s_ss' find the block from the visible letters, then fill each blank so that the block repeats without a break. Here the block is r s s s, and the blanks are s, r, s, r.

Letter grids. In a grid such as A D G / C F I / E H ?, treat every row as a series and every column as a series. The row step is +3 and the column step is +2, so the blank is K.