The number of natural numbers lying between 1012 and 23421 that can be formed using the digits $2,3,4,5,6$ (repetition of digits is not allowed) and divisible by 55 is _________.
Explanation:
When number is 4-digit number $(\overline{a b c d})$ here $d$ is fixed as 5
So, $(a, b, c)$ can be $(6,4,3),(3,4,6),(2,3,6)$, $(6,3,2),(3,2,4)$ or $(4,2,3)$
$\Rightarrow 6$ numbers
Case-II
No number possible
The number of matrices of order $3 \times 3$, whose entries are either 0 or 1 and the sum of all the entries is a prime number, is __________.
Explanation:
In a $3\times3$ order matrix there are $9$ entries.
These nine entries are zero or one.
The sum of positive prime entries are $2, 3, 5$ or $7$.
Total possible matrices $ = {{9!} \over {2!\,.\,7!}} + {{9!} \over {3!\,.\,6!}} + {{9!} \over {5!\,.\,4!}} + {{9!} \over {7!\,.\,2!}}$
$ = 34 + 84 + 126 + 36$
$ = 282$
A class contains b boys and g girls. If the number of ways of selecting 3 boys and 2 girls from the class is 168 , then $\mathrm{b}+3 \mathrm{~g}$ is equal to ____________.
Explanation:
${}^b{C_3}\,.\,{}^g{C_2} = 168$
$ \Rightarrow {{b(b - 1)(b - 2)} \over 6}\,.\,{{g(g - 1)} \over 2} = 168$
$ \Rightarrow b(b - 1)(b - 2)\,\,\,\,\,\,g(g - 1) = {2^5}{.3^2}.7$
$ \Rightarrow b(b - 1)(b - 2)\,\,\,\,\,\,g(g - 1) = 6\,.\,7\,.\,8\,.\,3\,.\,2$
$\therefore$ $b = 8$ and $g = 3$
$\therefore$ $b + 3g = 17$
Let $S$ be the set of all passwords which are six to eight characters long, where each character is either an alphabet from $\{A, B, C, D, E\}$ or a number from $\{1,2,3,4,5\}$ with the repetition of characters allowed. If the number of passwords in $S$ whose at least one character is a number from $\{1,2,3,4,5\}$ is $\alpha \times 5^{6}$, then $\alpha$ is equal to ___________.
Explanation:
If password is 6 character long, then
Total number of ways having atleast one number $ = {10^6} - {5^6}$
Similarly, if 7 character long $ = {10^7} - {5^7}$
and if 8-character long $ = {10^8} - {5^8}$
Number of password $ = ({10^6} + {10^7} + {10^8}) - ({5^6} + {5^7} + {5^8})$
$ = {5^6}({2^6} + {5.2^7} + {25.2^8} - 1 - 5 - 25)$
$ = {5^6}(64 + 640 + 6400 - 31)$
$ = 7073 \times {5^6}$
$\therefore$ $\alpha = 7073$.
Numbers are to be formed between 1000 and 3000 , which are divisible by 4 , using the digits $1,2,3,4,5$ and 6 without repetition of digits. Then the total number of such numbers is ____________.
Explanation:
Case-I
If first digit is 1
Then last two digits can be 24, 32, 36, 52, 56, 64
Case – II
If first digit is 2 then last two digit can be 16, 36, 56, 64
Total ways = 12 + 18 = 30 ways
The number of 5-digit natural numbers, such that the product of their digits is 36 , is __________.
Explanation:
Factors of 36 = 22 . 32 . 1
Five-digit combinations can be
(1, 2, 2, 3, 3) (1, 4, 3, 3, 1), (1, 9, 2, 2, 1)
(1, 4, 9, 11) (1, 2, 3, 6, 1) (1, 6, 6, 1, 1)
i.e., total numbers
${{5!} \over {2!2!}} + {{5!} \over {2!2!}} + {{5!} \over {2!2!}} + {{5!} \over {3!}} + {{5!} \over {2!}} + {{5!} \over {3!2!}}$
$ = (30 \times 3) + 20 + 60 + 10 = 180$.
The letters of the word 'MANKIND' are written in all possible orders and arranged in serial order as in an English dictionary. Then the serial number of the word 'MANKIND' is _____________.
Explanation:
Step 1 :
Write all the alphabets in alphabetical order.
Step 2 :
Give a number to each alphabet starting from 0 in alphabetical order.
Step 3 :
In word "MANKIND" first alphabet is "M" which is present in the 4th position in the Alphabet Box. And after "M", in the word "MANKIND" there are 6 alphabets "ANKIND" which can be arrange in ${{6!} \over {2!}}$ ways. Here $2!$ present because two N presents.
So for alphabet "M" we write $4.{{6!} \over {2!}}$
Step 4 :
As in word "MANKIND" we calculated "M", so delete "M" from alphabet box and check remaining alphabets got right numbers or not in alphabet box.
As you can see after deleting "M", 4th position in the alphabet box is missing so we create a new alphabet box.
Step 5 :
Now next alphabet after "M" in word "MANKIND" is "A" which is present in the 0th position in the new alphabet box. And after "A" in the word "MANKIND" there are 5 alphabets "NKIND" which can be arrange in ${{5!} \over {2!}}$ ways.
So, for alphabet "A" we write $0.{{5!} \over {2!}}$
Step 6 :
Now in word "MANKIND" we calculated "A", so delete "A" from previous new alphabet box and check remaining alphabets got right numbers or not in alphabet box.
As you can see after deleting "A", 0th positioin in the alphabet box is missing, so we have to create a new alphabet box.
Step 7 :
Now next alphabet after "A" in word "MANKIND" is "N" which is present in the 3rd and 4th position in the alphabet box. We have to choose minimum position for "N" in the alphabet box. So, we choose 3rd position "N". And after "N" in the word "MANKIND" there are 4 alphabets "KIND" which can be arrange ${{4!} \over {2!}}$ ways. Here $2!$ used because of 2 N's (one "N" is that "N" which we are considering and other "N" which is present in the word "KIND")
So for alphabet "N" we write $3\, \times \,{{4!} \over {2!}}$
Step 8 :
Now in word "MANKIND" we calculated "N", so delete 3rd "N" from alphabet box and check remaining alphabets got right number or not in alphabet box. As you can see after deleting 3rd "N", 3rd position is missing in the alphabet box so we have to create a new alphabet box.
Step 9 :
Now next alphabet after "N" in word "MANKIND" is "K" which is present at the 2nd position in the new alphabet box. And after "K" in the word "MANKIND" there are 3 alphabets "IND" which we can arrange $3!$ ways.
So for alphabet "K" we write $2\times3!$
Step 10 :
Now, in word "MANKIND" we calculated "K", so delete "K" from alphabet box and check remaining alphabets got right number or not in alphabet box. As you can see after deleting "K", 2nd position in the alphabet box is missing so we have to create a new alphabet box.
Step 11 :
Now next alphabet after "K" in word "MANKIND" is "I" which is present at 1st position in the new alphabet box. And after "I" in the word "MANKIND" there are 2 alphabets "ND" which we can arrange in $2!$ ways.
So for alphabet "I" we write $1\times2!$
Step 12 :
Now in word "MANKIND" we calculated "I", so delete "I" from alphabet box and check remaining alphabets got right numbers or not in alphabet box. As you can see after deleting "I", 1st position in the alphabet box is missing so we have to create a new alphabet box.
Step 13 :
Now next alphabet after "I" in word "MANKIND" is "N" which is present at 1st position in the new alphabet box. And after "N" in the word "MANKIND" there is 1 alphabet "D" which we can arrange in $1!$ ways.
So for alphabet "N" we write $1\times1!$.
Step 14 :
Now in word "MANKIND" we calculated "N", so delete "N" from alphabet box and check remaining alphabets got right numbering or not in alphabet box. As you can see after deleting "N", remaining alphabet "D" got correct numbering so new alphabet box will be same.
Step 15 :
Now next alphabet after "N" in word "MANKIND" is "D" which is present at the 0th position in the new alphabet box. And after "D" in the word "MANKIND" there is no alphabet so it is the last alphabet.
So for last alphabet we always write $0!$
$\therefore$ Position of the word "MANKIND" in the dictionary
$ = 4\,.\,{{6!} \over {2!}} + 0\,.\,{{5!} \over {2!}} + 3\,.\,{{4!} \over {2!}} + 2\,.\,3!\, + 1\,.\,2!\, + \,1\,.\,1!\, + \,0!$
$ = 1440 + 0 + 36 + 12 + 2 + 1 + 1$
$ = 1492$
The number of 6-digit numbers made by using the digits 1, 2, 3, 4, 5, 6, 7, without repetition and which are multiple of 15 is ____________.
Explanation:
A number is multiple of 15 when the number is divisible by 5 and sum of digits of the number is divisible by 3.
Among 1, 2, 3, 4, 5, 6, 7 unit place is filled with 5 so it is multiple of 5.
Now to make it divisible by 3, take remaining 5 digits such a way that sum becomes divisible by 3.
Remaining 5 digits can be
(1) 1, 2, 3, 4, 6
Here sum = 1 + 2 + 3 + 4 + 6 + 5 = 21 (divisible by 3)
This 5 digits can be filled in those 5 placed without repetition in 5 $\times$ 4 $\times$ 3 $\times$ 2 $\times$ 1 = 51 = 120 ways
(2) 2, 3, 4, 6, 7
Here sum = 2 + 3 + 4 + 6 + 7 + 5 = 27 (divisible by 3)
$\therefore$ Number of ways = 51 = 120
(3) 1, 2, 3, 6, 7
Here sum = 1 + 2 + 3 + 6 + 7 + 5 = 24 (divisible by 3)
$\therefore$ Number of ways = 51 = 120
$\therefore$ Total possible 6 digit numbers divisible by 15
= 120 + 120 + 120 = 360
The total number of four digit numbers such that each of first three digits is divisible by the last digit, is equal to ____________.
Explanation:
If unit digit is 2 then $\rightarrow 4 \times 5 \times 5=100$ numbers
If unit digit is 3 then $\rightarrow 3 \times 4 \times 4=48$ numbers
If unit digit is 4 then $\rightarrow 2 \times 3 \times 3=18$ numbers
If unit digit is 5 then $\rightarrow 1 \times 2 \times 2=4$ numbers
If unit digit is 6 then $\rightarrow 1 \times 2 \times 2=4$ numbers
For $7,8,9 \rightarrow 4+4+4=12$ Numbers
Total $=1086$ Numbers
Let b1b2b3b4 be a 4-element permutation with bi $\in$ {1, 2, 3, ........, 100} for 1 $\le$ i $\le$ 4 and bi $\ne$ bj for i $\ne$ j, such that either b1, b2, b3 are consecutive integers or b2, b3, b4 are consecutive integers. Then the number of such permutations b1b2b3b4 is equal to ____________.
Explanation:
There are 98 sets of three consecutive integer and 97 sets of four consecutive integers.
Using the principle of inclusion and exclusion,
Number of permutations of $b_{1} b_{2} b_{3} b_{4}=$ Number of permutations when $b_{1} b_{2} b_{3}$ are consecutive + Number of permutations when $b_{2} b_{3} b_{4}$ are consecutive - Number of permutations when $b_{1} b_{2}$ $b_{3} b_{4}$ are consecutive
$=97 \times 98+97 \times 98-97=97 \times 195=18915$.
Let A be a matrix of order 2 $\times$ 2, whose entries are from the set {0, 1, 2, 3, 4, 5}. If the sum of all the entries of A is a prime number p, 2 < p < 8, then the number of such matrices A is ___________.
Explanation:
$\because$ Sum of all entries of matrix A must be prime p such that 2 < p < 8 then sum of entries may be 3, 5 or 7.
If sum is 3 then possible entries are (0, 0, 0, 3), (0, 0, 1, 2) or (0, 1, 1, 1).
$\therefore$ Total number of matrices = 4 + 4 + 12 = 20
If sum of 5 then possible entries are
(0, 0, 0, 5), (0, 0, 1, 4), (0, 0, 2, 3), (0, 1, 1, 3), (0, 1, 2, 2) and (1, 1, 1, 2).
$\therefore$ Total number of matrices = 4 + 12 + 12 + 12 + 12 + 4 = 56
If sum is 7 then possible entries are
(0, 0, 2, 5), (0, 0, 3, 4), (0, 1, 1, 5), (0, 3, 3, 1), (0, 2, 2, 3), (1, 1, 1, 4), (1, 2, 2, 2), (1, 1, 2, 3) and (0, 1, 2, 4).
Total number of matrices with sum 7 = 104
$\therefore$ Total number of required matrices
= 20 + 56 + 104
= 180
The number of ways, 16 identical cubes, of which 11 are blue and rest are red, can be placed in a row so that between any two red cubes there should be at least 2 blue cubes, is _____________.
Explanation:
Here, $x_1+x_2+x_3+x_4+x_5+x_6=11$ and $x_2, x_3, x_4, x_5 \geq 2$
So $x_1+x_2+x_3+x_4+x_5+x_6=3$
No. of solutions $={ }^8 C_5=56$
The total number of 3-digit numbers, whose greatest common divisor with 36 is 2, is ___________.
Explanation:
$\because$ x $\in$ [100, 999], x $\in$ N
Then ${x \over 2}$ $\in$ [50, 499], ${x \over 2}$ $\in$ N
Number whose G.C.D. with 18 is 1 in this range have the required condition. There are 6 such number from 18 $\times$ 3 to 18 $\times$ 4. Similarly from 18 $\times$ 4 to 18 $\times$ 5 ......., 26 $\times$ 18 to 27 $\times$ 18
$\therefore$ Total numbers = 24 $\times$ 6 + 6 = 150
The extra numbers are 53, 487, 491, 493, 497 and 499.
There are ten boys B1, B2, ......., B10 and five girls G1, G2, ........, G5 in a class. Then the number of ways of forming a group consisting of three boys and three girls, if both B1 and B2 together should not be the members of a group, is ___________.
Explanation:
Number of ways when B1 and B2 are not together
= Total number of ways of selecting 3 boys $-$ B1 and B2 are together
= 10C3 $-$ 8C1
= ${{10\,.\,9\,.\,8} \over {1\,.\,2\,.\,3}} - 8$
= 112
Number of ways to select 3 girls = 5C3 = 10
$\therefore$ Total number of ways = 112 $\times$ 10 = 1120
The total number of three-digit numbers, with one digit repeated exactly two times, is ______________.
Explanation:
$C - 1:$ All digits are non-zero
${}^9{C_2}\,.\,2\,.\,{{3!} \over 2} = 216$
$C - 2$ : One digit is 0
$0,\,0,\,x \Rightarrow {}^9{C_1}\,.\,1 = 9$
$0,x,\,x \Rightarrow {}^9{C_1}\,.\,2 = 18$
Total $ = 216 + 27 = 243$
The number of 3-digit odd numbers, whose sum of digits is a multiple of 7, is _____________.
Explanation:
$\therefore$ x y 1, x y 3, x y 5, x y 7, x y 9 are the type of numbers. numbers.
If $x \,y\, 1$ then $x+y=6,13,20$... Cases are required
i.e., $6+6+0+\ldots=12$ ways
If $x \, y \,3$ then
$x+y=4,11,18, \ldots$ Cases are required
i.e., $4+8+1+0 \ldots=13$ ways
Similarly for $x \,y \,5$, we have $x+y=2,9,16, \ldots$
i.e., $2+9+3=14$ ways
for $x \,y$ we have
$x+y=0,7,14, \ldots$
i.e., $0+7+5=12$ ways
And for $x$ y we have
$ x+y=5,12,19 \ldots $
i.e., $5+7+0 \ldots=12$ ways
$\therefore$ Total 63 ways
Let A be a 3 $\times$ 3 matrix having entries from the set {$-$1, 0, 1}. The number of all such matrices A having sum of all the entries equal to 5, is ___________.
Explanation:
$\begin{aligned} 1 & \rightarrow 7 \text { times } \\\\ \text { and }-1 & \rightarrow 2 \text { times } \end{aligned}$
number of possible marrix $=\frac{9 !}{7 ! 2 !}=36$
Case-II:
$1 \rightarrow 6$ times,
$-1 \rightarrow 1$ times
and $0 \rightarrow 2$ times
number of possible marrix $=\frac{9 !}{6 ! 2 !}=252$
Case-III:
$ 1 \rightarrow 5$ times,
and $0 \rightarrow 4$ times
number of possible marrix $=\frac{9 !}{5 ! 4 !}=126$
Hence total number of all such matrix $A$ $=414$
The number of 7-digit numbers which are multiples of 11 and are formed using all the digits 1, 2, 3, 4, 5, 7 and 9 is _____________.
Explanation:
Multiple of $11 \rightarrow$ Difference of sum at even and odd place is divisible by 11 .
Let number of the form abcdefg
$ \begin{aligned} &\therefore(\mathrm{a}+\mathrm{c}+\mathrm{e}+\mathrm{g})-(\mathrm{b}+\mathrm{d}+\mathrm{f})=11 \mathrm{x} \\\\ &\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}+\mathrm{e}+\mathrm{f}=31 \\\\ &\therefore \text { either } \mathrm{a}+\mathrm{c}+\mathrm{e}+\mathrm{g}=21 \text { or } 10 \\\\ &\therefore \mathrm{b}+\mathrm{d}+\mathrm{f}=10 \text { or } 21 \end{aligned} $
Case-1
$ \begin{aligned} &a+c+e+g=21 \\\\ &b+d+f=10 \\\\ &(b, d, f) \in\{(1,2,7)(2,3,5)(1,4,5)\} \\\\ &(a, c, e, g) \in\{(1,4,7,9),(3,4,5,9),(2,3,7,9)\} \end{aligned} $
Case-2
$ \begin{aligned} &\mathrm{a}+\mathrm{c}+\mathrm{e}+\mathrm{g}=10 \\\\ &\mathrm{~b}+\mathrm{d}+\mathrm{f}=21 \\\\ &(\mathrm{a}, \mathrm{b}, \mathrm{e}, \mathrm{g}) \in\{1,2,3,4)\} \\\\ &(\mathrm{b}, \mathrm{d}, \mathrm{f}) \&\{(5,7,9)\} \\\\ &\therefore \text { Total number in case } 2=3 ! \times 4 !=144 \\\\ &\therefore \text { Total numbers }=144+432=576 \end{aligned} $
In an examination, there are 5 multiple choice questions with 3 choices, out of which exactly one is correct. There are 3 marks for each correct answer, $-$2 marks for each wrong answer and 0 mark if the question is not attempted. Then, the number of ways a student appearing in the examination gets 5 marks is ____________.
Explanation:
$3 x-2 y=5$ and $x+y \leq 5$ where $x, y \in \mathrm{W}$
Only possible solution is $(x, y)=(3,2)$
Students can mark correct answers by only one choice but for an incorrect answer, there are two choices. So total number of ways of scoring 5 marks $={ }^{5} C_{3}(1)^{3} \cdot(2)^{2}=40$
Explanation:
FARMER (6)
A, E, F, M, R, R
| A | |||||
|---|---|---|---|---|---|
| E | |||||
| F | A | E | |||
| F | A | M | |||
| F | A | R | E | ||
| F | A | R | M | E | R |
Explanation:
When 4 consonants are together (V, W, L, S)
such cases = 3! ⋅ 4! = 144
All consonants should not be together
= Total $-$ All consonants together,
= 6! $-$ 3! 4! = 576
Explanation:
3n $-$ 1 type $\to$ 2, 5 = Q
3n $-$ 2 type $\to$ 1, 4 = R
number of subset of S containing one element which are not divisible by 3 = ${}^2$C1 + ${}^2$C1 = 4
number of subset of S containing two numbers whose some is not divisible by 3
= ${}^3$C1 $\times$ ${}^2$C1 + ${}^3$C1 $\times$ ${}^2$C1 + ${}^2$C2 + ${}^2$C2 = 14
number of subsets containing 3 elements whose sum is not divisible by 3
= ${}^3$C2 $\times$ ${}^4$C1 + (${}^2$C2 $\times$ ${}^2$C1)2 + ${}^3$C1(${}^2$C2 + ${}^2$C2) = 22
number of subsets containing 4 elements whose sum is not divisible by 3
= ${}^3$C3 $\times$ ${}^4$C1 + ${}^3$C2(${}^2$C2 + ${}^2$C2) + (${}^3$C1${}^2$C1 $\times$ ${}^2$C2)2
= 4 + 6 + 12 = 22
number of subsets of S containing 5 elements whose sum is not divisible by 3.
= ${}^3$C3(${}^2$C2 + ${}^2$C2) + (${}^3$C2${}^2$C1 $\times$ ${}^2$C2) $\times$ 2 = 2 + 12 = 14
number of subsets of S containing 6 elements whose sum is not divisible by 3 = 4
$\Rightarrow$ Total subsets of Set A whose sum of digits is not divisible by 3 = 4 + 14 + 22 + 22 + 14 + 4 = 80.
Explanation:
| 5 | a | b | b | a | 5 |
For divisible by 55 it shall be divisible by 11 and 5 both, for divisibility by 5 unit digit shall be 0 or 5 but as the number is six digit palindrome unit digit is 5.
A number is divisible by 11 if the difference between sum of the digits in the odd places and the sum of the digits in the even places is a multiple of 11 or zero.
Sum of the digits in the even place = a + b + 5
Sum of the digits in the odd places = a + b + 5
Difference between the two sums = (a + b + 5 ) - (a + b + 5) = 0
0 is divisible by 11.
Hence, 5abba5 is divisible by 11.
So, required number = 10 $\times$ 10 = 100
Explanation:
= 1! + 2 . 2! + 3 . 3! + ..... 15 $\times$ 15!
$ = \sum\limits_{r = 1}^{15} {(r + 1)! - (r)!} $
= 16! $-$ 1
= ${}^{16}{P_{16}}$ $-$ 1
$\Rightarrow$ q = r = 16, s = 1
${}^{q + s}{C_{r - s}} = {}^{17}{C_{15}}$ = 136
Explanation:
Number of numbers = 20
(ii) When 4 or 6 are at unit place
Number of numbers = 32
Total three digit even number = 20 + 32 = 52
Explanation:
Now, power of 2 must be zero,
power of 5 can be anything,
power of 13 can be anything
But, power of 11 should be even.
So, required number of divisors is
1 $\times$ 11 $\times$ 14 $\times$ 6 = 924
Explanation:
Total student $\matrix{ 5 & 6 & 8 \cr } $
$\matrix{ 2 & 3 & 5 \cr } \Rightarrow $ ${}^5{C_2} \times {}^6{C_3} \times {}^8{C_5}$
Number of selection $\matrix{ 2 & 2 & 6 \cr } \Rightarrow {}^5{C_2} \times {}^6{C_2} \times {}^8{C_6}$
$\matrix{ 3 & 2 & 5 \cr } \Rightarrow {}^5{C_3} \times {}^6{C_2} \times {}^8{C_5}$
$\Rightarrow$ Total number of ways = 23800
According to question 100 K = 23800
$\Rightarrow$ K = 238
Explanation:
= 4 $\times$ 4 $\times$ 3 $\times$ 2 = 96
Explanation:
6 : Bowlers
7 : Batsman
2 : Wicket keepers
Total number of ways for :
at least 4 bowler, 5 batsman & 1 wicket keeper
= ${}^6{C_4}({}^7{C_6} \times {}^2{C_1} + {}^7{C_5} \times {}^2{C_2}) + {}^6{C_5} \times {}^7{C_5} \times {}^2{C_1}$
$ = 777$
Explanation:
$\sum\limits_{r = 1}^{10} {r![(r + 1)(r + 2)(r + 3) - 9(r + 1) + 8]} $
$ = \sum\limits_{r = 1}^{10} {[\{ (r + 3)! - (r + 1)!\} - 8\{ (r + 1)! - r!\} ]} $
$ = (13! + 12! - 2! - 3!) - 8(11! - 1)$
$ = (12\,.\,13 + 12 - 8)\,.\,11! - 8 + 8 = (160)(11!)$
Therefore, $\alpha = 160$
Explanation:
In double digit numbers = 10 + 9 = 19
In triple digit numbers = 100 + 90 + 90 = 280
Total = 300 times
Explanation:
$=($Greater number - Smaller number)(Greater number - Smaller number)!
i.e. $1=(2-1)^{(2-1) !}, $
$4^{24}=(12-8)^{(12-8) !}, $
$3^6=(7-4)^{(7-4) !}$
$\therefore \quad ?=(5-3)^{(5-3) !}$
$\therefore$ Required number $=2^{2 !}=2^{2 \times 1}=4$
Explanation:
The possible combination of 3 digits numbers are
1, 2, 3; 1, 2, 4; 1, 2, 5; 1, 3, 4; 1, 3, 5; 1, 4, 5; 2, 3, 4; 2, 3, 5; 2, 4, 5; and 3, 4, 5.
The possible combination of numbers which are divisible by 3 are 1, 2, 3; 3, 4, 5; 1, 3, 5 and 2, 3, 4.
(If sum of digits of a number is divisible by 3 then the number is divisible by 3)
$ \therefore $ Total number of numbers = 4 × 3! = 24
The possible combination of numbers divisible by 5 are 1, 2, 5; 2, 3, 5; 3, 4, 5; 1, 3, 5; 1, 4, 5 and 2, 4, 5.
(If the last digit of a number is 0 or 5 then the number is divisible by 5)
$ \therefore $ Total number of numbers = 6 × 2! = 12
The possible combination of number divisible by both 3 and 5 are 1, 3, 5 and 3, 4, 5.
$ \therefore $ Total number of numbers = 2 $ \times $ 2! = 4
$ \therefore $ Total required number = 24 + 12 - 4 = 32
Explanation:
= 10C1 [29 – 2] = 5100
If group C has two students then number of groups
= 10C2 [28 – 2] = 11430
If group C has three students then number of groups
= 10C3 × [27 – 2] = 15120
So total groups = 31650
Explanation:
Vowels $ \to $ EE
Total No of words = ${{6!} \over {2!2!}}$ = 180
Total no of words if vowels are together
= ${{5!} \over {2!}}$ = 60
$ \therefore $ Total no of words where
vowels never come together = 180 – 60 = 120.
Explanation:
$P = {2 \over 6} = {1 \over 3}$
$ \therefore $ $q = 1 - {1 \over 3} = {2 \over 3}$ (not showing 3 or 5)
Experiment is performed with 4 dices independently
$ \therefore $ Their binomial distribution is
${(q + p)^4} = {(q)^4} + {}^4{C_1}{q^3}p + {}^4{C_2}{q^2}{p^2} + {}^4{C_3}q{p^3} + {}^4{C_4}{P^4}$
$ \therefore $ In one throw of each dice probability of showing 3 or 5 at least twice is
= ${p^4} + {}^4{C_3}q{p^3} + {}^4{C_2}{q^2}{p^2}$
$ = {{33} \over {81}}$
Given such experiment performed 27 times
$ \therefore $ So expected outcomes = np
= ${{33} \over {81}} \times 27$
= 11
Explanation:
1. Two S letters
2. Two L letters
3. One Y letter
4. One A letter
5. One B letter
6. One U letter
Number of ways we can select two alike letters = 2C1
Then number of ways we can select two distinct letters = 5C2
Then total arrangement of selected letters = ${{4!} \over {2!}}$
So total number of words, with or without meaning, that can be formed
= 2C1 $ \times $ 5C2 $ \times $ ${{4!} \over {2!}}$ = 240
Explanation:
Answering right option for each question is possible in 1 way.
So ways of choosing right option for 4 questions = 1.1.1.1 = (1)4
Number of ways of choosing wrong option for each question = 3
So ways of choosing wrong option for 2 questions = (3)2
$ \therefore $ Required number of ways = 6C4.(1)4.(3)2 = 135
Explanation:
Given that sum of digits = 10
$ \therefore $ x + y + z = 10 ......(1)
Also x can't be 0 as if x = 0 then it will become 2 digits number.
So, x $ \ge $ 1, y $ \ge $ 0, z $ \ge $ 0
As x $ \ge $ 1
$ \Rightarrow $ x $-$ 1 $ \ge $ 0
Let x $-$ 1 = t
$ \therefore $ t $ \ge $ 0
From equation (1)
(x $-$ 1) + y + z = 9
$ \Rightarrow $ t + y + z = 9
Now this problem becomes, distributing 9 things among 3 people t, y, z.
Number of ways we can do that
= ${}^{9 + 3 - 1}{C_{3 - 1}} = {}^{11}{C_2} = 55$
Now when 3 digit number is 900 then t = 9, y = 0, z = 0.
And when t = 9, then
x $-$ 1 = 9
$ \Rightarrow $ x = 10
But we can't take x = 10 in a 3 digits number. So, we have to remove this case.
$ \therefore $ Total number of 3 digit numbers = 55 $-$ 1 = 54.
Explanation:
Explanation:
To form four letter words
Case 1 : All same ( not possible)
Case 2 : 1 different, 3 same (not possible)
Case 3 : 2 different, 2 same
= 3C1 $ \times $ 7C2 $ \times $ ${{4!} \over {2!}}$ = 756
Case 4 : 2 same of one kind, 2 same same of other kind
= 3C2 $ \times $ ${{4!} \over {2!2!}}$ = 18
Case 5 : All letters are different
= 8C4 $ \times $ 4! = 1680
$ \therefore $ Total ways = 1680 + 756 + 18 = 2454
Explanation:
No of ways 4 marbels can be chosen where atmost 3 red marbels can be present.
Case 1: When 3 red marbels present
No of ways = 5C3 $ \times $ 7C1
Case 2: When 2 red marbels present
No of ways = 5C2 $ \times $ 7C2
Case 3: When 1 red marbels present
No of ways = 5C1 $ \times $ 7C3
Case 4: When 0 red marbels present
No of ways = 5C0 $ \times $ 7C4
$ \therefore $ Total number of ways
= 5C3 $ \times $ 7C1 + 5C2 $ \times $ 7C2 + 5C1 $ \times $ 7C3 + 5C0 $ \times $ 7C4
= 70 + 210 + 175 + 35
= 490
Let $\mathrm{S}=\{1,2,3,4,5,6,7,8,9\}$. Let $x$ be the number of 9-digit numbers formed using the digits of the set S such that only one digit is repeated and it is repeated exactly twice. Let $y$ be the number of 9 -digit numbers formed using the digits of the set S such that only two digits are repeated and each of these is repeated exactly twice. Then,
$56 x=9 y$
$21 x=4 y$
$45 x=7 y$
$29 x=5 y$
The letters of the word "UDAYPUR" are written in all possible ways with or without meaning and these words are arranged as in a dictionary. The rank of the word "UDAYPUR" is
1579
1578
1580
1581
The largest value of $n$, for which $40^n$ divides $60!$, is
14
13
11
12
The number of ways, in which 16 oranges can be distributed to four children such that each child gets at least one orange, is
384
403
429
455
The largest $n \in \mathbb{N}$, for which $7^n$ divides $101!$, is :
18
15
19
16
The number of strictly increasing functions $f$ from the set $\{1,2,3,4,5,6\}$ to the set $\{1,2,3, \ldots ., 9\}$ such that $f(i) \neq i$ for $1 \leq i \leq 6$, is equal to :
21
28
27
22
There are 12 points in a plane, no three of which are in the same straight line, except 5 points which are collinear. Then the total number of triangles that can be formed with the vertices at any three of these 12 points is
230
210
200
220
From a group of 7 batsmen and 6 bowlers, 10 players are to be chosen for a team, which should include atleast 4 batsmen and atleast 4 bowlers. One batsmen and one bowler who are captain and vice-captain respectively of the team should be included. Then the total number of ways such a selection can be made, is