Given: The digits \[2,\text{ }3,\text{ }0,\text{ }3,\text{ }4,\text{ }2,\text{ }3\] Total number of digits \[=\text{ }7\] We know, zero cannot be the first digit of the \[7\] digit numbers. Number...

Given: The word \[MADHUBANI\] Total number of letters \[=\text{ }9\] A total number of arrangements of word \[MADHUBANI\] excluding \[I\] Total letters \[8\] Repeating letter \[A\]  repeating twice....

Given: The word \[SERIES\] There are \[6\] letters in the word \[SERIES\] out of which \[2\] are \[Ss,\text{ }2\text{ }are\text{ }Es\] and the rest all are distinct. Now, Let us fix \[5\] letters at...

Given: The digits \[0,\text{ }1,\text{ }1,\text{ }5,\text{ }9\] Total number of digits \[=\text{ }5\] So now, number greater than \[50000\] will have either \[5\text{ }or\text{ }9\] in the first...

There are \[7\] letters in the word \[ARRANGE\]out of which \[2\text{ }are\text{ }As,\text{ }2\text{ }are\text{ }Rs\] and the rest all are distinct. So by using the formula, \[n!/\text{ }\left(...

Given: The digits \[1,\text{ }3,\text{ }3,\text{ }0\] Total number of digits \[=\text{ }4\] Digits of the same type \[=\text{ }2\] Total number of \[4\]digit numbers \[=\text{ }4!\text{ }/\text{...

Given: Number of red flags \[=\text{ }4\] Number of white flags \[=\text{ }2\] Number of green flags \[=\text{ }3\] So there are total \[9\] flags, out of which \[4\] are red, \[2\] are white,...

Given: The digits \[1,\text{ }2,\text{ }3,\text{ }4,\text{ }3,\text{ }2,\text{ }1\] The total number of digits are \[7\] There are \[4\] odd digits \[1,1,3,3\text{ }and\text{ }4\]odd places \[\left(...

Given: The word \[MUMBAI\] There are \[6\] letters in the word \[MUMBAI\]out of which \[2\] are \[Ms\] and the rest all are distinct. So let us consider both \[Ms\]together as one letter, the...

Given: The word \[PARALLEL\] There are \[8\] letters in the word \[PARALLEL\] out of which \[2\text{ }are\text{ }As,\text{ }3\text{ }are\text{ }Ls\] and the rest all are distinct. So by using the...

There are \[9\text{ }\left( i.e\text{ }powers\text{ }3\text{ }+\text{ }2\text{ }+\text{ }4\text{ }=\text{ }9 \right)\]objects in the expression \[{{a}^{3}}~{{b}^{2}}~{{c}^{4}}\] and there are...

Given: The word \[UNIVERSITY\] There are \[10\] letters in the word \[UNIVERSITY\] out of which \[2\text{ }are\text{ }Is\] There are \[4\]vowels in the word \[UNIVERSITY\] out of which \[2\text{...

Given: The word \[ALGEBRA\] There are \[4\]consonants in the word \[ALGEBRA\] The number of ways to arrange these consonants is \[^{4}{{P}_{4}}~=\text{ }4!\] There are \[3\] vowels in the word...

\[CONSTANTINOPLE\] There are \[14\] letters in the word \[CONSTANTINOPLE\]out of which \[2\text{ }are\text{ }Os,\text{ }3\text{ }are\text{ }Ns,\text{ }2\text{ }are\text{ }Ts\]and the rest all are...

(i) \[SERIES\] There are \[6\]letters in the word \[SERIES\]out of which \[2\text{ }are\text{ }Ss,\text{ }2\text{ }are\text{ }Es\]and the rest all are distinct. So by using the formula, \[n!/\text{...

(i) \[PAKISTAN\] There are \[8\]letters in the word \[PAKISTAN\]out of which \[2\text{ }are\text{ }As\]and the rest all are distinct. So by using the formula, \[n!/\text{ }\left( p!\text{ }\times...

(i) \[ARRANGE\] There are \[7\] letters in the word \[ARRANGE\] out of which\[2\text{ }are\text{ }As,\text{ }2\text{ }are\text{ }Rs\] and the rest all are distinct. So by using the formula,...

(i) \[INDEPENDENCE\] There are \[12\] letters in the word \[INDEPENDENCE\] out of which \[2\text{ }are\text{ }Ds,\text{ }3\text{ }are\text{ }Ns,\text{ }4\text{ }are\text{ }Es\]and the rest all are...

Solution: (iii) Are the vowels always together? There are 4 vowels and 6 consonants in the word ‘GANESHPURI’. Consider 4 (A,E,I,U) vowels as one letter, then total number of letters is 7 (A,E,I,U,...

Solution: The word 'GANESHPURI' is made up of ten letters. 10P10 = 10! is the total amount of words created! (i) the letter G always occupies the first place? If the first position is fixed with the...

Solution: Total number of letters in the word ‘SUNDAY’ is 6. So, number of arrangements of 6 things, taken all at a time is 6P6 = 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720 Now, we shall find the number of...

Solution: Given: In the word 'ORIENTAL,' there are four vowels (O, I, E, A) The number of consonants in the given word is equal to four (R, N, T, L) Positions that are unusual include (1, 3, 5 or 7)...

Solution: The word 'SUNDAY' has a total of 6 letters. So 6P6 is the number of configurations of 6 objects taken all at once. = 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720 Now we'll count the number of words...

Solution: (iii) the vowels occupy only the odd places? The word 'STRANGE' has seven letters. The vowels are made up of these letters (A,E). In the word 'STRANGE,' there are four odd positions. The...

Solution: The word 'STRANGE' has seven letters, including two vowels (A,E) and five consonants (S,T,R,N,G). (i) the vowels come together? If we consider two vowels to be one letter, we'll end up...

Solution: A school's total number of teachers is 36. We know that P is equal to the number of configurations of n things taken r at a time (n, r) Using the formula, you can $ P\text{ }\left(...

Solution: P equals the number of arrangements of 'n' objects taken all at once (n, n) As a result, by employing the formula, $ P\text{ }\left( n,\text{ }r \right)\text{ }=\text{ }n!/\left( n-r...

Solution: Given: n+5Pn+1 = 11(n – 1)/2 n+3Pn P (n +5, n + 1) = 11(n – 1)/2 P(n + 3, n) By using the formula, P (n, r) = n!/(n – r)! (n + 5) (n + 4) = 22 (n – 1) n2 + 4n + 5n + 20 = 22n – 22 n2 + 9n...

Solution: It is given that $ P\left( 15,\text{ }r-1 \right)\text{ }:\text{ }P\left( 16,\text{ }r-2 \right)\text{ }=\text{ }3\text{ }:\text{ }4 $ $ P\left( 15,\text{ }r-1 \right)\text{ }/\text{...

Solution: By making use of the formula, $ P\text{ }\left( n,\text{ }r \right)\text{ }=\text{ }n!/\left( n-r \right)! $ $ P\text{ }\left( n,\text{ }n \right)\text{ }=\text{ }n!/\left( n-n \right)! $...

Solution: It is given that: $ P\left( n,\text{ }5 \right)\text{ }:\text{ }P\left( n,\text{ }3 \right)\text{ }=\text{ }2\text{ }:\text{ }1 $ $ P\left( n,\text{ }5 \right)\text{ }/\text{ }P\left(...

Solution: Given: P(2n – 1, n) : P(2n + 1, n – 1) = 22 : 7 P(2n – 1, n) / P(2n + 1, n – 1) = 22 / 7 By using the formula, P (n, r) = n!/(n – r)! P (2n – 1, n) = (2n – 1)! / (2n – 1 – n)! = (2n – 1)!...

Solution: It is given that: $ P\text{ }\left( n-1,\text{ }3 \right):\text{ }P\text{ }\left( n,\text{ }4 \right)\text{ }=\text{ }1\text{ }:\text{ }9 $ $ P\text{ }\left( n-1,\text{ }3 \right)/\text{...

Solution: It is given that P (n, 4) = 12. P (n, 2) By making use of the formula, $ P\text{ }\left( n,\text{ }r \right)\text{ }=\text{ }n!/\left( n-r \right)! $ $ P\text{ }\left( n,\text{ }4...

Solution: It is given that : P (11, r) = P (12, r – 1) By making use of the formula, $ P\text{ }\left( n,\text{ }r \right)\text{ }=\text{ }n!/\left( n-r \right)! $ $ P\text{ }\left( 11,\text{ }r...

Solution: Given: P (9, r) = 3024 By using the formula, P (n, r) = n!/(n – r)! P (9, r) = 9!/(9 – r)! So, from the question, P (9, r) = 3024 Substituting the obtained values in above expression we...

Solution: It is given that nP4 = 360 We can write nP4 as P (n , 4) By making use of the formula, P (n, r) = n!/(n – r)! P (n, 4) = n!/(n – 4)! So, according to the question, nP4 = P (n, 4) = 360...

Solution: It is given that P(n, 5) = 20 P(n, 3) By making use of the formula, & P\text{ }\left( n,\text{ }r \right)\text{ }=\text{ }n!/\left( n-r \right)! \\ & P\text{ }\left( n,\text{ }5...

Solution: It is given that => 5 P(4, n) = 6 P(5, n – 1) By making use of the formula, => P (n, r) = n!/(n – r)! => P (4, n) = 4!/(4 – n)! => P (5, n-1) = 5!/(5 – (n-1))! = 5!/(5 – n +...

Solution: According to the question, P (5, r) = P (6, r – 1) By making use of the formula, $ P\text{ }\left( n,\text{ }r \right)\text{ }=\text{ }n!/\left( n-r \right)! $ $ P\text{ }\left( 5,\text{...

(iii) 6P6 (iv) P (6, 4) Solution: (iii) 6P6 We can write 6P6 as P (6, 6) By making use of the formula, P (n, r) = n!/(n – r)! We get: P (6, 6) = 6!/(6 – 6)! 6P6 = 6!/0! 6P6 = (6 × 5 × 4 × 3 × 2 ×...

(i) 8P3 (ii) 10P4 Solution: (i) 8P3 We can write that 8P3 as P (8, 3) By making use of the formula of permutation, we get: $ P\text{ }\left( n,\text{ }r \right)\text{ }=n!/\left( n-r \right)! $ $...

Solution: Numbers in excess of 8000 are required. As a result, two numerals can be used in the thousand place: 8 or 9. Assume four boxes; the first box can contain one of the two numbers 8 or 9,...

Solution: In a nine-digit number, there can't be a 0 in the initial digit, and digits can't be repeated. As a result, there are 9C1= 9 different methods to fill in the initial digit. There are now 9...

Solution: Assume there are six seats available. Any one of six members can sit in the first seat, bringing the total number of options to 6C1. The second seat can be occupied by any of the five...

Solution: The required number of people is over 7000. As a result, any of the three numerals in the thousand's place can be used: 7, 8, 9. Assume four boxes; the first box can contain one of the...

Solution: (i) We know that the initial digit of the license plates cannot be zero. It is also forbidden to repeat digits. Let's say there are five boxes. The first box can be filled with any of the...

Solution: We know that the last digit of an odd number is (1, 3, 5, 7, 9). Assume there are three boxes. Any of the nine digits can be entered in the first box (zero not allowed at first position)...

Solution: Let us assume we have three boxes. The first box can be filled with any one of the nine digits (zero not allowed at first position) So, possibilities are 9C1 The second box can be filled...

Solution: Assume there are three boxes. Any of the nine digits can be entered in the first box (0 not allowed at first place). As a result, the alternatives are numerous. 9C1 Any of the nine digits...

Solution: We understand that each AP has a distinct initial word and a common difference. As a result, the number of possibilities to choose the first term of a set is 3C1 = 3 And there are 5C1 = 5...

 Solution: A race between twelve students is held. The number of methods to choose the first prize winner is 12C1 There are 11C1 different ways to choose the winner of the second reward (11, since...

Solution: One squad contains 6 boys and 4 girls, while the other has 5 boys and 3 girls. Assume that team 1 consists of 6 boys and 4 girls. Team 2 will consist of 5 guys and 3 girls. Singles bouts...

Solution: There are seven flags available, of which two are required to make a signal. From this, we can deduce that we must choose two flags from a total of seven and arrange them in order to...

Solution: (i) Given: there are five mathematics books and six physics books. The number of ways to purchase one mathematics book is  5C1. Similarly, there are 6C1 methods to purchase one physics...

Solution: Only one of the given alternatives is correct in a multiple-choice question. Only one of the four answers to the first three questions is correct. As a result, there are four possible...

Solution: The total number of possible approaches to unlocking the lock is =  10 × 10 × 10 = 1000 The number of times the lock has been opened successfully = 1. The number of unsuccessful...

Solution: According to the question, there are two options for an examinee's response to a question: true or false. An examinee can answer a series of ten true/false questions in any of the...

Solution: According to the question, a coin is tossed five times, with the outcome being either heads or tails each time, leaving two choices. The following are all possible outcomes:...

Solution: According to the question, the total number of packets is four and there are 5 post offices in all. One parcel can be posted in five different ways, including at any of the two post...

Solution: According to the question, Mint has to perform (i) Choose the number of days in February (28 or 29 days are possible), and (ii) Choose February 1st as the first day of the month. Now, Mint...

Solution: According to the question, there are two routes from Goa to Bombay which is air and sea. So, the number of ways in which a person can travel from Goa to Bombay is given by 2C1 According to...

Solution: According to the question, there are 10 fountain pens, 12 ball pens, and 5 pencils and the person has to (i) choose a ball pen from 12 ball pens. (ii) choose a fountain pen from 10...

Solution: According to the question, there are 27 boys and 14 girls and the teacher has to (i) choose a boy from 27 boys, and (ii) choose a girl from 14 girls. The number of ways in which a boy can...

Solution: It is given that: n! (n + 2) = n! + (n + 1)! Let us take the RHS first and solve accordingly! $ n!\text{ }+\text{ }\left( n\text{ }+\text{ }1 \right)!\text{ }=\text{ }n!\text{ }+\text{...

Solution: (i) (2 + 3)! = 2! + 3! Let us consider LHS: (2 + 3)! (2 + 3)! = 5! Now RHS, 2! + 3! = (2×1) + (3×2×1) = 2 + 6 = 8 LHS ≠ RHS ∴ The given expression is false. (ii) (2 × 3)! = 2! × 3! Let us...

Solution: (iii) (n + 1) (n + 2) (n + 3) … (2n) We have: $ \left( n\text{ }+\text{ }1 \right)\text{ }\left( n\text{ }+\text{ }2 \right)\text{ }\left( n+3 \right)\ldots \left( 2n \right)=\left[ \left(...

Solution: (i) 5 ⋅ 6 ⋅ 7 ⋅ 8 ⋅ 9 ⋅ 10 We can write the above equation as: $ 5~\cdot ~6~\cdot ~7~\cdot ~8~\cdot ~9~\cdot ~10=\left( 1\times 2\times 3\times 4\times 5\times 6\times 7\times 8\times...

Solution: (i) 1/4! + 1/5! = x/6! We can write 4! and 5! as: 5! = 5 × 4 × 3 × 2 × 1 6! = 6 × 5 × 4 × 3 × 2 × 1 So by making use of these values, $ 1/4!\text{ }+\text{ }1/5!\text{ }=\text{ }x/6! $ $...

Solution: We are given that: 1/9! + 1/10! + 1/11! = 122/11! First, let us consider the LHS: $ 1/9!\text{ }+\text{ }1/10!\text{ }+\text{ }1/11!\text{ }=\text{ }1/9!\text{ }+\text{ }1/\left( 10\times...

Solution: We have to find the LCM of (6!, 7!, 8!) We know that we can write: 8! = 8 × 7 × 6! 7! = 7 × 6! So, making use of the above values we get: $ L.C.M.\text{ }\left( 6!,\text{ }7!,\text{ }8!...

Solution: (i) 30!/28! Using the properties of factorials! $ 30!/28!=\left( 30\times 29\times 28! \right)/28! $ $ 30!/28!=30\times 29 $ $ 30!/28!=870 $ (ii) (11! – 10!)/9! Using the properties of...