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Finding the Rank of a Word

To find the rank of a word out of all the possibilities using all the letters given in the word is nothing but the extension of the concept of alphabetically arranging the words in a dictionary. However, unlike the case of the dictionary, we can have ‘meaningless’ words also in the case of finding the rank.
 
Example-1
All the letters of the word ‘LUCKNOW’ are arranged in all possible ways. What will be the rank of the word LUCKNOW?
Solution
Alphabetical order of occurrence of letters— C, K, L, N, O, U, W.
 
Number of words starting with C = 6!
 
Number of words starting with K = 6!
 
All the words starting with LC – 5!
 
All the words starting with LK – 5!
 
All the words starting with LN – 5!
 
All the words starting with LO – 5!
 
Next word will start with LU – C – K – N – O – W.
 
So, rank of LUCKNOW – 2 × 6! + 4 × 5! + 1 = 1921.
 
 
Example-2
In a chess competition involving some boys and girls of a school, every student had to play exactly one game with every other student. It was found that in 45 games both the players were girls, and in 190 games both were boys. The number of games in which one player was a boy and the other was a girl is
  1. 200
  2. 216
  3. 235
  4. 256
Solution
Let there be m boys and n girls
nC2 = 45 = Description: 2333.png⇒ n(n − 1) = 90 ⇒ n = 10
mC2 = 190 = Description: 2337.png = 190 ⇒ m (m − 1) = 380 ⇒ m = 20
 
Number of games between one boy and one girl = 10C1 × 20C1 = 10 × 20 = 200
 
Hence, option (a).
 
 
Example-3
If each permutation of the digits 1, 2, 3, 4, 5, 6 is listed in the increasing order of the magnitude, the 289th term will be
  1. 326541
  2. 341256
  3. 356241
  4. 314256
Solution
289 = (2×5!) + (2×4!) + 1
 
So, the number will be 341256.
 
 
Example-4
There are 12 intermediate stations between two places A and B. In how many ways can a train be made to stop at 4 of these 12 intermediate stations provided no two of them are consecutive?
Solution
1st Method Let S1, S2, …, S8 denote the stations where the train does not stop. The four stations where the train stops should be at any four of the nine places indicated by cross.
∴ Required number = Description: 2341.png
 
2nd Method Let S1, S2, S3, S4 be the four intermediate stations where the train stops.
Description: 2350.png
 
Let a, b, c, d, e be the number of stations between A and S1, S1 and S2, S2 and S3, S3 and S4 and B respectively.
 
Then, a + b + c + d + e = 8…(1)
 
Where a ≥ 0, b ≥ 1, c ≥ 1, d ≥ 1, e ≥ 0
 
Let x = ay = b – 1, z = c – 1, t = d – 1, w = e
 
Now x + y + z + t + w a + b + c + d + e – 3 = 8 – 3 = 5
 
Or x + y + z + t + w = 5, where x, y, z, t, w ≥ 0…(2)
 
Required number = number of non negative integral solutions
n+r-1Cr 5+5-1C59C5 = 126
 
 
Example-5
Find the number of integral solutions of equation x + y + z + t = 25, x > 0, y > 1, z > 2 and t ≥ 0.
Solution
Given, x + y + z + t = 25,…(1)
 
Where x ≥ 1, y ≥ 2, z ≥ 3, t ≥ 0
 
Let p x – 1, q = y – 2, r = z – 3, s = t
 
Then p + q + r + s = x + y + z + t – 6 = 25 – 6 = 19, where, p, q, r, s ≥ 0
 
∴ p + q + r + s = 19, p, q, r, s ≥ 0…(2)
 
∴ Required number = number of ways in which 19 identical things can be distributed among 4 persons when each person can get any number of things = n+r-1Cr-1 = 22C3
 
 
Example-6
There are 4 oranges, 5 apricots and 6 alphonso in a fruit basket. In how many ways can a person make a selection of fruits from among the fruits in the basket?
Solution
Whenever we are talking about fruits, we assume them to be identical. However when we are talking about men, we treat them to be distinct.
 
Zero or more oranges can be selected out of 4 identical oranges in 4 + 1 = 5 ways.
 
Zero or more apricot can be selected out of 5 identical apricots in 5 +1 = 6 ways.
 
Zero or more can be selected out of 6 identical alphanso in 6 + 1 = 7 ways.
 
∴ The total number of selections when all the three types of fruits are selected (the number of any type of fruit may also be zero)
= 5 × 6 × 7 = 210.
 
But in one of these selections the number of each type of fruit is zero and hence this selection must be excluded.
∴ Required number = 210 – 1 = 209.
 
 
Example-7
Twelve different letters of alphabet are given. Words with six letters are formed from these given letters. Find the number of words which have at least one letter repeated.
Solution
The total number of letters = 12. Words of six letters are to be framed.
 
The total number of words of 6 letters when any letter may be repeated any number of times (This also includes the number of words formed when no letter is repeated.)
 
= 12 × 12 × 12 × 12 × 12 × 12 = 126
 
Number of words of 6 letters when no letter is repeated = 12P6.
 
So, Number of words of 6 letters which have at least one letter repeated = 126 – 12P6
 




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