How many words can be formed from the letters of the word ‘SERIES’,
Question: How many words can be formed from the letters of the word SERIES, which start with S and end with S? Solution: To find: number of words which start and end with S There are 4 places to fill up with 4 letters out of which 2 are of the same kind $\Rightarrow$ Number of words $=\frac{4 !}{2 !}=12$ 12 words are possible...
Read More →Find the number of arrangements of the letters of the word ‘ALGEBRA’
Question: Find the number of arrangements of the letters of the word ALGEBRA without altering the relative positions of the vowels and the consonants. Solution: To find: number of arrangements without changing the relative position The following table shows where the vowels and consonants can be placed Consonants can be placed in the blank places There are 3 spaces for vowels There are 3 vowels out of which 2 are alike Vowels can be placed in $\frac{3 !}{2 !}=3$ ways There are 4 consonants, and ...
Read More →How many words can be formed by arranging the letters of the word ‘INDIA’,
Question: How many words can be formed by arranging the letters of the word INDIA, so that the vowels are never together? Solution: To find: Number of words that can be formed so that vowels are never together Number of words such that vowels are never the together $=$ Total number of words Number of words where vowels are together Total number of words $=\frac{5 !}{2 !}=60$ To find a number of words where vowels are together Let the vowels I, I, A be represented by a single letter Z $\Rightarro...
Read More →How many words can be formed by arranging the letters of the word
Question: How many words can be formed by arranging the letters of the word ARRANGEMENT, so that the vowels remain together? Solution: To find: number of words where vowels are together Vowels in the above word are: $A, A, E, E$ Consonants in the above word: R,R,N,G,M,N,T Let us denote the all the vowels by a single letter say $Z$ $\Rightarrow$ The word now has the letters, R,R,N,G,M,N,T,Z $\mathrm{R}$ and $\mathrm{N}$ are repeated twice Number of permutations $=\frac{8 !}{2 ! 2 !}$ Now $Z$ is c...
Read More →How many different signals can be transmitted by arranging
Question: How many different signals can be transmitted by arranging 2 red, 3 yellow and 2 green flags on a pole, if all the seven flags are used to transmit a signal? Solution: To find: Number of distinct signals possible Total number of fags = 7 2 are of 1 kind, 3 are of another kind, and 2 are of the 3rd kind $\Rightarrow$ Number of distinct signals $=\frac{7 !}{2 ! 3 ! 2 !}=210$ Hence 210 different signals can be made...
Read More →A child has three plastic toys bearing the digits 3, 3, 5 respectively.
Question: A child has three plastic toys bearing the digits 3, 3, 5 respectively. How many 3 - digit numbers can he make using them? Solution: To find: number of 3 digit numbers he can make If all were distinct, he could have made $3 !=6$ numbers But 2 number are the same So the number of possibilities $=\frac{3 !}{2 !}=\frac{6}{2}=3$ He can make 3 three - digit numbers using them...
Read More →There are 3 blue balls, 4 red balls and 5 green balls.
Question: There are 3 blue balls, 4 red balls and 5 green balls. In how many ways can they are arranged in a row? Solution: To find: no of ways in which the balls can be arranged in a row where some balls are of the same kind Total number of balls $=3+4+5=12$ 3 are of 1 kind, 4 are of another kind, 5 are of the third kind Number of ways $=\frac{12 !}{3 ! 4 ! 5 !}=27720$ They can be arranged in 27720 ways...
Read More →In how many ways can the letters of the expression
Question: In how many ways can the letters of the expression $x^{2} y^{2} z^{4}$ be arranged when written without using exponents? Solution: To find: number of ways the letters can be arranged The following table shows the possible arrangements However, we see that case $1=x^{2} y^{2} z^{4}$ is the same as case $4=y^{2} x^{2} z^{4}$ Similarly (case2, case 5), (case 3, case 6) are the same So there are only 3 distinct cases Hence the letters can be arranged in 3 distinct ways...
Read More →Find the total number of permutations of the letters of each of the words
Question: Find the total number of permutations of the letters of each of the words given below: (i) APPLE (ii) ARRANGE (iii) COMMERCE (iv) INSTITUTE (v) ENGINEERING (vi) INTERMEDIATE Solution: To find: number of permutations of the letters of each word Number of permutations of n distinct letters is n! Number of permutations of n letters where r letters are of one kind, s letters of another kind, $t$ letters of a third kind and so on $=\frac{n !}{r ! s ! t ! \ldots}$ (i) Here $n=5$ P is repeate...
Read More →Find the number of ways in which m boys and n girls may be arranged in a
Question: Find the number of ways in which m boys and n girls may be arranged in a row so that no two of the girls are together; it is given that m n. Solution: In this question, n girls are to be seated alternatively between m boys. There are m+1 spaces in which girls can be arranged. The number of ways of arranging $n$ girls is $P(m+1, n)=\frac{(m+1) !}{(m-n+1) !}$ ways. Formula: Number of permutations of $n$ distinct objects among $r$ different places, where repetition is not allowed, is $P(n...
Read More →when a group photograph is taken, all the seven teachers should be in the
Question: when a group photograph is taken, all the seven teachers should be in the first row, and all the twenty students should be in the second row. If the tow corners of the second row are reserved for the two tallest students, interchangeable only between them, and if the middle seat of the front row is reserved for the principal, how many arrangements are possible? Solution: For the first row: There are 7 teachers in which the position of principal is fixed. Therefore, the teachers can be ...
Read More →In how many ways can 5 children be arranged in a line such that
Question: In how many ways can 5 children be arranged in a line such that (i) two of them, Rajan and Tanvy, are always together? (ii) two of them, Rajan and Tanvy, are never together, Solution: (i) two of them, Rajan and Tanvy, are always together Consider Rajan and Tanvy as a group which can be arranged in $2 !=2$ ways. The 3 children with this 1 group can be arranged in $4 !=24$ ways. The total number of possibilities in which they both come together is $2 \times 24=48$ ways. (ii) Two of them,...
Read More →In an examination, there are 8 candidates out of which 3 candidates have to
Question: In an examination, there are 8 candidates out of which 3 candidates have to appear in mathematics and the rest in different subjects. In how many ways can they are seated in a row if candidates appearing in mathematics are not to sit together? Solution: Candidates in mathematics are not sitting together = total ways the Students are appearing for mathematic sit together. The total number of arrangements of 8 students is $8 !=40320$ When students giving mathematics exam sit together, th...
Read More →How many numbers divisible by 5 and lying between 3000 and 4000 can be
Question: How many numbers divisible by 5 and lying between 3000 and 4000 can be formed by using the digits 3, 4, 5, 6, 7, 8 when no digit is repeated in any such number? Solution: For a number to be divisible by 5, the last digit should either be 5 or 0. In this case, 5 is only possible. For a four digit number to be between 3000 to 4000 , in this case, should start with 3 . Therefore, the other 2 digits can be arranged by 4 numbers in $P(4,2)$ Formula: Number of permutations of $n$ distinct ob...
Read More →How many permutations can be formed by the letters of the word ‘VOWELS’, when
Question: How many permutations can be formed by the letters of the word VOWELS, when (i) there is no restriction on letters; (ii) each word begins with E; (iii) each word begins with $\mathrm{O}$ and ends with $\mathrm{L}$; (iv) all vowels come together; (v) all consonants come together? Solution: (i) There is no restriction on letters The word VOWELS contain 6 letters. The permutation of letters of the word will be $6 !=720$ words. (ii) Each word begins with Here the position of letter $E$ is ...
Read More →Find the number of ways in which the letters of the word ‘MACHINE’
Question: Find the number of ways in which the letters of the word MACHINE can be arranged such that the vowels may occupy only odd positions. Solution: To find: number of words Condition: vowels occupy odd positions. There are 7 letters in the word MACHINE out of which there are 3 vowels namely A C E. There are 4 odd places in which 3 vowels are to be arranged which can be done $P(4,3)$. The rest letters can be arranged in $4 !$ ways Formula: Number of permutations of $n$ distinct objects among...
Read More →In how many arrangements of the word ‘GOLDEN’ will the vowels never
Question: In how many arrangements of the word GOLDEN will the vowels never occur together? Solution: To find: number of words Condition: vowels should never occur together. There are 6 letters in the word GOLDEN in which there are 2 vowels. Total number of words in which vowels never come together = Total number of words - total number of words in which the vowels come together. A total number of words is $6 !=720$ words. Consider the vowels as a group. Hence there are 5 groups that can be arra...
Read More →In how many ways can the letters of the word ‘FAILURE’ be arranged so
Question: In how many ways can the letters of the word FAILURE be arranged so that the consonants may occupy only odd positions? Solution: To find: number of words Condition: consonants occupy odd places There are total of 7 letters in the word FAILURE. There are 3 consonants, i.e. $F, L, R$ which are to be arranged in 4 places. The rest 5 letters can be arranged in 4! Ways. Formula: Number of permutations of n distinct objects among $r$ different places, where repetition is not allowed, is $P(n...
Read More →How many words can be formed out of the letters of the word ‘ORIENTAL’
Question: How many words can be formed out of the letters of the word ORIENTAL so that the vowels always occupy the odd places? Solution: To find: number of words formed Condition: vowels occupy odd places There are 8 letters in the word ORIENTAL and vowels are 4 which are O, I, E,A respectively. There is 4 odd places in which 4 vowels are to be arranged. The rest 4 letters can be arranged in $4 !$ Ways. Formula: Number of permutations of $n$ distinct objects among $r$ different places, where re...
Read More →In how many ways can the letters of the word ‘HEXAGON’ be permuted?
Question: In how many ways can the letters of the word HEXAGON be permuted? In how many words will the vowels be together? Solution: There are 7 letters in the word HEXAGON. Formula: Number of permutations of $n$ distinct objects among $r$ different places, where repetition is not allowed, is $P(n, r)=n ! /(n-r) !$ Therefore, a permutation of 7 different objects in 7 places is $P(7,7)=\frac{\frac{7 !}{(7-7) !}}{=}=\frac{7 !}{0 !}=\frac{5040}{1}=5040$ They can be permuted in P (7,7) = 5040 ways. ...
Read More →Find the number of permutations of the letters of the word ‘ENGLISH’.
Question: Find the number of permutations of the letters of the word ENGLISH. How many of these begin with E and end with I? Solution: There are 7 letters in the word ENGLISH. Permutation of 7 letters in 7 places can be done in $\mathrm{P}(7,7)$ ways. Formula: Number of permutations of n distinct objects among r different places, where repetition is not allowed, is $P(n, r)=n ! /(n-r) !$ Therefore, a permutation of 7 different objects in 7 places is $P(7,7)=\frac{7 !}{(7-7) !}=\frac{7 !}{0 !}=\f...
Read More →How many words beginning with C and ending with Y can be formed by
Question: How many words beginning with C and ending with Y can be formed by using the letters of the word COURTESY? Solution: To find: number of words starting with C and end with Y There are 8 letters in word COURTESY. Here the position of the letters $C$ and $Y$ are fixed which is $1^{\text {st }}$ and $8^{\text {th }}$. Rest 6 letters are to be arranged in 6 places which can be done in $P(6,6)$. Formula: Number of permutations of $n$ distinct objects among $r$ different places, where repetit...
Read More →How many words can be formed from the letters of the word ‘SUNDAY’?
Question: How many words can be formed from the letters of the word SUNDAY? How many of these begin with D? Solution: There are 6 letters in the word SUNDAY. Different words formed using 6 letters of the word SUNDAY is $P(6,6)$ Formula: Number of permutations of $n$ distinct objects among $r$ different places, where repetition is not allowed, is $P(n, r)=n ! /(n-r) !$ Therefore, a permutation of 6 different objects in 6 places is $P(6,6)=\frac{6 !}{(6-6) !}=\frac{6 !}{0 !}=\frac{720}{1}=720$ 720...
Read More →Find the number of different 4-letter words (may be meaningless) that can
Question: Find the number of different 4-letter words (may be meaningless) that can be formed from the letters of the word NUMBERS, Solution: To find: 4 lettered word from letters of word NUMBERS There are 7 alphabets in word NUMBERS. The word is a 4 different letter word. Formula: Number of permutations of $n$ distinct objects among $r$ different places, where repetition is not allowed, is $P(n, r)=n ! /(n-r) !$ Therefore, a permutation of 7 different objects in 4 places is $P(7,4)=\frac{7 !}{(...
Read More →Find the number of words formed (may be meaningless) by using all the
Question: Find the number of words formed (may be meaningless) by using all the letters of the word EQUATION, using each letter exactly once. Solution: There are 8 alphabets in the word EQUATION. Formula: Number of permutations of $n$ distinct objects among $r$ different places, where repetition is not allowed, is $P(n, r)=n ! /(n-r) !$ Therefore, a permutation of 8 different objects in 8 places is $P(8,8)=\frac{8 !}{(8-8) !}=\frac{8 !}{0 !}=\frac{40320}{1}=40320$ Hence there are 40320 words for...
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