Solve the Following Question.(5 Marks)

Question 15 Marks

If the letters of the word 'MOTHER' are written in all possible orders and these words are written out as in a dictionary, find the rank of the word 'MOTHER'.

Answer

In the dictionary the words at each stage are arranged in alphabetical order. In the given problem we must therefore consider the words beginning with E, H, M, 0, R, T in order. E will occur in the first place as often as there are ways of arranging the remaining 5 letters,
$\therefore$ Number of words starting with E = 5! = 5 × 4 × 3 × 2 × 1 = 120
Number of words starting with H = 5! = 120.
Number of words beginning with M is 5!, but one of these words is the word MOTHER.
So, we first find the number of words beginning with ME and MH.
Number of words starting with ME = 4! = 4 × 3 × 2 × 1 = 24.
Now, the words beginning with 'MO' must follow.
There are 4! words beginning with MO, one of these words is the word MOTHER itself.
So, we first find the number of words beginning with MOE, MOH and MOR.
Number of words starting with MOE = 3! = 6
Number of words starting with MOH = 3! = 6
Number of words starting with MOR = 3! = 6
Number of words beginning with MOT is 3! but one of these words is the word MOTHER itself
So, we first find the number of words beginning with MOTE.
Number of words starting with MOTE = 2! = 2
Now, the words beginning with MOTH must follow.
There are 2! words beginning with MOTH, one of these words is word MOTHER itself.
The first word beginning with MOTH is the word MOTHER.
$\therefore$ Rank of MOTHER = 2 × 120 + 2 × 24 + 3 × 6 + 2 + 1
= 240 + 48 + 18 + 3
= 309

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Question 25 Marks

The letters of the word 'ZENITH' are written in all possible orders. How many words are possible if all these words are written out as in a dictionary? What is the rank of the word 'ZENITH'?

Answer

In a dictionary the words at each stage are arranged in alphabetical order. In the given problem we must therefore consider the words beginning with E, H, I, N, T, Z in order. 'E' will occur in the first place as often as there are ways of arranging the remaining 5 letters all at a time i.e. E will occur S! times. Similarly H will occur in the first place the same number of times. $\therefore$ Number of words starting with E = 5! = 5 × 4 × 3 × 2 × 1 = 120Number of words starting with H = 5! = 120
Number of words starting with I = 5! = 120
Number of words starting with N = 5! = 120
Number of words starting with T = 5! = 120
Number of words beginning with Z is S!, but one of these words is the word ZEN ITH i tselr. So, we first find the number of words beginning with ZEH, ZEI and ZENH,
Number of words starting with ZEH = 3! = 6
Number of words starting with ZEI = 3! = 6
Number of words starting with ZENH = 2! = 2.
Now, the words beginning with ZENI must follow.
There are 21 words beginning with ZENI one of these words is the word ZENITH itself. The first word beginning with ZENI is the word ZENI HT and the next word is ZENITH.
$\therefore$ Rank of ZENITH = 5 × 120 + 2 × 6 + 2 + 2
= 600 + 12 + 4
= 600 + 16
= 616

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Question 35 Marks

In how many ways can the letters of the word "INTERMEDIATE" be arranged so that:

The vowels always occupy even places?
The relative order of vowels and consonants do not alter?

Answer

INTERMEDIATEI = 2 times, T = 2 times, E = 3 times, N, R, M, D, A
Number of letters = 12

There are 6 vowels. They ocuepy even places 2nd, 4th, 6th, 8th, 10th, 12th. After there six there are six places and 5 letters, T is 2 times. So, number of ways for consonants $=\frac{6!}{2!}$

The total number of ways when vowels ocuepy even places
$=\frac{6!}{2!}\times\frac{6!}{2!\ 3!}$
$=\frac{6\times5\times4\times3\times2\times6\times5\times4\times3\times2}{2\times2\times3\times2}$
$=21600$
Required number of ways= 21600

Number of ways such that relative order of vowels and consonants do not alter

$=\frac{6!}{2!}\times\frac{6!}{2!\ 3!}$
$=21600$
Required number of ways= 21600

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Question 45 Marks

If the permutations of a, b, c, d, e taken all together be written down in alphabetical order as in dictionary and numbered, find the rank of the permutation debac.

Answer

In a dictionary the words at each stage are arranged in alphabetical order. In the given problem we must therefore consider the words beginning with a, b, c, d, e in order. 'a' will occur in the first place as often as there are ways of arranging the remaining 4 letters all at a time i.e 'a' will occur 4! times. similarly b and c will occur in the first place the same number of times,
$\therefore$ Number of words starting with'a' = 4! = 4 × 3 × 2 × 1 = 24
Number of words starting with 'b' = 4! = 4 × 3 × 2 × 1 = 24
Number of words starting with 'c' = 4! = 4 × 3 × 2 × 1 = 24
Number of words beginning with 'd' is 4!, but one of these words is the word debac. So, we first find the number of words beginning with da, db, de, and dea
Number of words starting with da = 3! = 6
Number of words starting with db = 3! = 6
Number of words starting with de = 3! = 6
Number of words starting with dea = 2! = 2
There are 2! words beginning with deb one of these words is the word debac itself .
The first word beginning with deb is the word debac.
$\therefore$ Rank of debac = 3 × 24 + 3 × 6 + 2 + 1
= 72 + 18 + 3
= 90 + 3
= 93

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Maths Solve the Following Question.(5 Marks)

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