Permutations and Combinations
Three persons enter in a lift at the ground floor. The lift will go up to 10th floor. The number of ways, in which the three persons can exit the lift at three different floors, if the lift does not stop at first, second and third floors, is equal to ________.
Explanation:
Given:
The lift goes up to the $10^{\text {th }}$ floor.
The lift does not stop at $1^{\text {st }}, 2^{\text {nd }}, 3^{\text {rd }}$ floors.
So, the persons can get down only at the remaining floors.
$ 4,5,6,7,8,9,10 \Rightarrow 7 \text { floors } $
Now, all three persons must exit at three different floors (no two can get down at the same floor).
Also, the three persons are distinct, so who gets down where matters.
First, choose which 3 floors (out of 7) will be used:
$ { }^7 C_3=\frac{7 \cdot 6 \cdot 5}{{3} \cdot 2 \cdot 1}=35 $
Next, assign (arrange) the 3 distinct persons to these 3 chosen floors:
Number of arrangements of 3 persons:
$ 3!=6 $
So, total number of ways:
$ \text { Total ways }={ }^7 C_3 \times 3!=35 \times 6=210 $
Explanation:
To find numbers of number greater than 5000 and less than 9000 so number will start with 5 or 9 of the form 5 abc and $a, b, c \in\{0,1,2,5,9\}$ also the number is divisible by 3 .
This means $5+\mathrm{a}+\mathrm{b}+\mathrm{c}$ is divisible by 3
so the sum $\mathrm{a}+\mathrm{b}+\mathrm{c}$ is of the type $3 \mathrm{n}+1$.
unit place c determined by the sum of previous 3 places.
rem(a $+b)=$ remainder of $a+b$
| rem(a+b) | combination (a,b) | value of c | total = combination (a, b) × c(choice) |
|---|---|---|---|
| 0 | (0, 0), (0, 9), (9, 0), (9, 9), (1, 2), (2, 1), (5, 1), (1, 5) | 1 | 8 × 1 = 8 |
| 1 | (0, 1), (9, 1), (1, 0), (1, 9), (2, 2), (2, 5), (5, 2), (5, 5) | 0, 9 | 8 × 2 = 16 |
| 2 | (0, 2), (0, 5), (9, 2), (9, 5), (2, 0), (2, 9), (5, 0), (5, 9), (1, 1) | 2, 5 | 9 × 2 = 18 |
Total = $8+16+18=42$
Let S denote the set of 4-digit numbers $a b c d$ such that $a>b>c>d$ and P denote the set of 5 -digit numbers having product of its digits equal to 20 . Then $n(\mathrm{~S})+n(\mathrm{P})$ is equal to $\_\_\_\_$
Explanation:
for set $S$ : 4-digit numbers $a b c d$ with $a>b>c>d$
digits are chosen from $\{0,1,2, \ldots, 9\}$
since order is strictly fixed, any 4 distinct digits chosen will form exactly one such number. number of ways to choose 4 digits from 10 is ${ }^{10} C_4$
$ n(S)=\frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1}=210 $
for set $P$ : 5-digit numbers with product of digits $=20$
possible sets of digits (using digits 1-9):
case 1: $\{5,4,1,1,1\}$
number of arrangements $=\frac{5!}{3!}=20$
case 2: $\{5,2,2,1,1\}$
number of arrangements $=\frac{5!}{2!2!}=30$
$ n(P)=20+30=50 $
calculating final value.
$ n(S)+n(P)=210+50=260 $
the value of $n(S)+n(P)$ is 260 .
The number of 4 -letter words, with or without meaning, which can be formed using the letters PQRPQRSTUVP, is $\_\_\_\_$ .
Explanation:
Letter frequency
$ \mathrm{P}: 3 $
$\mathrm{R}, \mathrm{Q}: 2$
$S, T, U, V: 1$
4 letter words can be of type
$\Rightarrow \mathrm{ABCD}$ or $\mathrm{AABC}, \mathrm{AABB}, \mathrm{AAAB}$
$ \begin{aligned} & \Rightarrow{ }^7 C_4 \cdot 4!+\left({ }^3 C_1\right) \cdot{ }^6 C_2 \cdot \frac{4!}{2!}+{ }^3 C_2 \cdot \frac{4!}{2!1!}+\left({ }^1 C_1\right) \cdot{ }^6 C_1 \cdot \frac{4!}{3!} \\ & =1422 \end{aligned} $
Let ABC be a triangle. Consider four points $\mathrm{p}_1, \mathrm{p}_2, \mathrm{p}_3, \mathrm{p}_4$ on the side AB , five points $p_5, p_6, p_7, p_8, p_9$ on the side $B C$, and four points $p_{10}, p_{11}, p_{12}, p_{13}$ on the side AC . None of these points is a vertex of the triangle ABC . Then the total number of pentagons, that can be formed by taking all the vertices from the points $p_1, p_2, \ldots, p_{13}$, is $\_\_\_\_$
Explanation:
$ { }^4 C_2 \cdot{ }^5 C_2 \cdot{ }^4 C_1+{ }^4 C_2 \cdot{ }^5 C_2 \cdot{ }^4 C_1+{ }^4 C_2 \cdot{ }^4 C_2 \cdot{ }^5 C_1=240+240+180=660 $
Let $S=\{(m, n): m, n \in\{1,2,3, \ldots . ., 50\}\}$. If the number of elements $(m, n)$ in $S$ such that $6^m+9^n$ is a multiple of 5 is $p$ and the number of elements ( $m, n$ ) in $S$ such that $m+n$ is a square of a prime number is q , then $\mathrm{p}+\mathrm{q}$ is equal to $\_\_\_\_$ .
Explanation:
$ \begin{aligned} & 5 \mid 6^m+9^n \\ & \Rightarrow 5 \mid 1^m+(-1)^n \end{aligned} $
$\Rightarrow \quad m$ and $n$ has to be opposite parity.
$ { }^2 C_1 \times{ }^{25} C_1 \cdot{ }^{25} C_1=625 \times 2=1250 $
For, $m+n=K^2$ for some prime $K$.
$ \begin{aligned} & m+n \in\{2,3, \ldots, 100\} \\ & m+n=4,9,25,49 \end{aligned} $
$m+n=4 \Rightarrow 3$ pairs
$m+n=9 \Rightarrow 8$ pairs
$m+n=25 \Rightarrow 24$ pairs
$m+n=49 \Rightarrow 48$ pairs
$\Rightarrow 83$ pairs
$\Rightarrow 1333$
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
Let m and $\mathrm{n},(\mathrm{m}<\mathrm{n})$, be two 2-digit numbers. Then the total numbers of pairs $(\mathrm{m}, \mathrm{n})$, such that $\operatorname{gcd}(m, n)=6$, is __________ .
Explanation:
$\begin{aligned} & m=6 a, n=6 b \\ & \text { So } \operatorname{gcd}(m, n)=6 \Rightarrow \operatorname{gcd}(a, b)=1 \\ & m=6 a \geq 10 \Rightarrow a \geq\left[\frac{10}{6}\right]=2 \\ & m=6 a \leq 99 \Rightarrow a \leq\left[\frac{99}{6}\right]=16 \end{aligned}$
So $a, b \in\{2,3, \ldots, 16\}$, and we count how many coprime pairs $(a, b)$ with $a< b, \operatorname{gcd}(a, b)=1$
$\begin{aligned} & a=2 \Rightarrow b=3,5,7,9,11,13,15 \Rightarrow 7 \\ & a=3 \Rightarrow \mathrm{~b}=4,5,7,8,10,11,13,14,16 \Rightarrow 9 \\ & a=4 \Rightarrow b=5,7,9,11,13,15 \Rightarrow 6 \\ & a=5 \Rightarrow \mathrm{~b}=6,7,8,9,11,12,13,14,16 \Rightarrow 9 \\ & a=6 \Rightarrow \mathrm{~b}=7,11,13 \Rightarrow 3 \\ & a=7 \Rightarrow b=8,9,10,11,12,13,15,16 \Rightarrow 8 \\ & a=8 \Rightarrow \mathrm{~b}=9,11,13,15 \Rightarrow 4 \\ & a=9 \Rightarrow \mathrm{~b}=10,11,13,14,16 \Rightarrow 5 \\ & a=10 \Rightarrow b=11,13 \Rightarrow 2 \\ & a=11 \Rightarrow b=12,13,14,15,16 \Rightarrow 5 \\ & a=12 \Rightarrow b=13,17 \times \rightarrow \text { only } 13 \text { is valid } \Rightarrow 1 \\ & a=13 \Rightarrow b=14,15,16 \Rightarrow 3 \\ & a=14 \Rightarrow b=15, \Rightarrow 1 \\ & a=15 \Rightarrow b=16 \Rightarrow 1 \end{aligned}$
$\begin{aligned} & \text { Total }=7+9+6+9+3+8+4+5+2+5+1 \\ & +3+1+1=64\end{aligned}$
All five letter words are made using all the letters A, B, C, D, E and arranged as in an English dictionary with serial numbers. Let the word at serial number $n$ be denoted by $\mathrm{W}_{\mathrm{n}}$. Let the probability $\mathrm{P}\left(\mathrm{W}_{\mathrm{n}}\right)$ of choosing the word $\mathrm{W}_{\mathrm{n}}$ satisfy $\mathrm{P}\left(\mathrm{W}_{\mathrm{n}}\right)=2 \mathrm{P}\left(\mathrm{W}_{\mathrm{n}-1}\right), \mathrm{n}>1$.
If $\mathrm{P}(\mathrm{CDBEA})=\frac{2^\alpha}{2^\beta-1}, \alpha, \beta \in \mathbb{N}$, then $\alpha+\beta$ is equal to :____________
Explanation:
Firstly, by this rule, we note:
$ P(W_1) = p $
$ P(W_2) = 2p $
$ P(W_3) = 4p $
…
$ P(W_n) = 2^{n-1}p $
To find the initial probability $ p $, consider the total probability must sum to 1 across all 120 possible words (since $ 5! = 120 $):
$ \sum_{n=1}^{120} P(W_n) = 1 $
This is a geometric series sum where:
$ p(1 + 2 + 2^2 + \ldots + 2^{119}) = 1 $
Since the series sum $ 1 + 2 + 2^2 + \ldots + 2^{119} $ is equal to $ 2^{120} - 1 $, we have:
$ p(2^{120} - 1) = 1 \Rightarrow p = \frac{1}{2^{120} - 1} $
Thus, the probability for the $ n $-th word is:
$ P(W_n) = \frac{2^{n-1}}{2^{120} - 1} \quad \text{(i)} $
Next, determine the position of "CDBEA". Starting from the first letter:
Words beginning with 'A': $ 4! = 24 $
Words beginning with 'B': $ 4! = 24 $
Words beginning with 'C':
CA*: $ 3! = 6 $
CB*: $ 3! = 6 $
CDA$ **: 2! = 2 $
CDBA*: $ 1! = 1 $
Summing these, the position of "CDBEA" is the 64th word.
Substitute into equation (i):
$ P(\text{CDBEA}) = P(W_{64}) = \frac{2^{63}}{2^{120} - 1} $
Given $ P(\text{CDBEA}) = \frac{2^\alpha}{2^\beta - 1} $, we find:
$ \alpha = 63 $
$ \beta = 120 $
Thus, the sum $ \alpha + \beta = 63 + 120 = 183 $.
If the number of seven-digit numbers, such that the sum of their digits is even, is $m \cdot n \cdot 10^n ; m, n \in\{1,2,3, \ldots, 9\}$, then $m+n$ is equal to__________
Explanation:
When numbers are uniformly distributed, half of them have even digit sums and half have odd digits sums.
Number of 7-digit numbers with even digit sum =
$\frac{1}{2} \cdot 9 \cdot 10^6=4.5 \cdot 10^6$
Note that $9 \cdot 5 \cdot 10^5=4.5 \cdot 10^6$
$m+n=9+5=14$
The number of 6-letter words, with or without meaning, that can be formed using the letters of the word MATHS such that any letter that appears in the word must appear at least twice, is _________.
Explanation:
(i) Single letter is used, then no. of words $=5$
(ii) Two distinct letters are used, then no. of words
${ }^5 \mathrm{C}_2 \times\left(\frac{6!}{2!4!} \times 2+\frac{6!}{3!3!}\right)=10(30+20)=500$
(iii) Three distinct letters are used, then no. of words
${ }^5 \mathrm{C}_3 \times \frac{6!}{2!2!2!}=900$
Total no. of words $=1405$
The number of natural numbers, between 212 and 999, such that the sum of their digits is 15, is _______.
Explanation:
$\begin{array}{|c|c|c|} \hline \mathrm{x} & \mathrm{y} & \mathrm{z} \\ \hline \end{array}$
Let $\mathrm{x}=2 \Rightarrow \mathrm{y}+\mathrm{z}=13$
$(4,9),(5,8),(6,7),(7,6),(8,5),(9,4), \rightarrow 6$
Let $x=3 \rightarrow y+z=12$
$(3,9),(4,8), \ldots \ldots . .,(9,3) \rightarrow 7$
Let $x=4 \rightarrow y+z=11$
$(2,9),(3,8), \ldots \ldots \ldots,(9,1) \rightarrow 9$
Let $x=5 \rightarrow y+z=10$
$(1,9),(2,8), \ldots \ldots . .,(9,1) \rightarrow 10$
Let $x=6 \rightarrow y+z=9$
$(0,9),(1,8), \ldots \ldots . .,(9,0) \rightarrow 9$
Let $\mathrm{x}=7 \rightarrow \mathrm{y}+\mathrm{z}=8$
$(0,9),(1,7), \ldots \ldots . .,(8,0) \rightarrow 9$
Let $x=8 \rightarrow y+z=7$
$(0,7),(1,6), \ldots \ldots . .,(7,0) \rightarrow 8$
Let $x=9 \rightarrow y+z=6$
$(0,6),(1,5), \ldots \ldots \ldots,(6,0) \rightarrow 7$
Total $=6=7+8+9+10+9+8+7=64$
Number of functions $f:\{1,2, \ldots, 100\} \rightarrow\{0,1\}$, that assign 1 to exactly one of the positive integers less than or equal to 98 , is equal to ________.
Explanation:
392 Ans.
The number of 3 -digit numbers, that are divisible by 2 and 3 , but not divisible by 4 and 9 , is _________.
Explanation:
No, of 3 digits $=999-99=900$
No. of 3 digit numbers divisible by 2 & 3 i.e. by 6
$\frac{900}{6}=150$
No. of 3 digit numbers divisible by 4 & 9 i.e. by 36
$\frac{900}{36}=25$
$\therefore$ No of 3 digit numbers divisible by $2 \& 3$ but not by $4 \& 9$
$150-25=125$
The number of ways, 5 boys and 4 girls can sit in a row so that either all the boys sit together or no two boys sit together, is ________.
Explanation:
A : number of ways that all boys sit together $=5!\times 5!$
B : number of ways if no 2 boys
sit together $=4!\times 5!$
$\mathrm{A} \cap \mathrm{B}=\phi$
Required no. of ways $=5!\times 5!+4!\times 5!=17280$
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
The number of sequences of ten terms, whose terms are either 0 or 1 or 2 , that contain exactly five 1 s and exactly three 2 s , is equal to :
If all the words with or without meaning made using all the letters of the word "KANPUR" are arranged as in a dictionary, then the word at 440th position in this arrangement is :
PRNAKU
PRKAUN
PRKANU
PRNAUK
Let $ P $ be the set of seven digit numbers with sum of their digits equal to 11. If the numbers in $ P $ are formed by using the digits 1, 2 and 3 only, then the number of elements in the set $ P $ is :
164
158
161
173
Let ${ }^n C_{r-1}=28,{ }^n C_r=56$ and ${ }^n C_{r+1}=70$. Let $A(4 \operatorname{cost}, 4 \sin t), B(2 \sin t,-2 \cos t)$ and $C\left(3 r-n, r^2-n-1\right)$ be the vertices of a triangle $A B C$, where $t$ is a parameter. If $(3 x-1)^2+(3 y)^2$ $=\alpha$, is the locus of the centroid of triangle ABC , then $\alpha$ equals
The number of different 5 digit numbers greater than 50000 that can be formed using the digits 0 , $1,2,3,4,5,6,7$, such that the sum of their first and last digits should not be more than 8 , is
Group A consists of 7 boys and 3 girls, while group B consists of 6 boys and 5 girls. The number of ways, 4 boys and 4 girls can be invited for a picnic if 5 of them must be from group $A$ and the remaining 3 from group $B$, is equal to :
The number of words, which can be formed using all the letters of the word "DAUGHTER", so that all the vowels never come together, is :
In a group of 3 girls and 4 boys, there are two boys $B_1$ and $B_2$. The number of ways, in which these girls and boys can stand in a queue such that all the girls stand together, all the boys stand together, but $B_1$ and $B_2$ are not adjacent to each other, is :
From all the English alphabets, five letters are chosen and are arranged in alphabetical order. The total number of ways, in which the middle letter is ' M ', is :
The number of integers, between 100 and 1000 having the sum of their digits equals to 14 , is __________.
Explanation:
Number in this range will be 3-digit number.
$N=\overline{a b c}$ such that $a+b+c=14$
Also, $a \geq 1, \quad a, b, c \in\{0,1,2, \ldots 9\}$
Case I
All 3-digit same
$\Rightarrow 3 a=14$ not possible
Case II
Exactly 2 digit same:
$\Rightarrow 2 a+c=14$
$\begin{aligned} & (a, c) \in\{(3,8),(4,6),(5,4),(6,2),(7,0)\} \\ & \Rightarrow\left(\frac{3!}{2!}\right) \text { ways } \Rightarrow 5 \times 3-1 \\ & =15-1=14 \end{aligned}$
Case III
All digits are distinct
$a+b+c=14$
without losing generality $a > b > c$
$\begin{aligned} & (a, b, c) \in\left\{\begin{array}{l} (9,5,0),(9,4,1),(9,3,2) \\ (8,6,0),(8,5,1),(8,4,2) \\ (7,6,1),(7,5,2),(7,4,3) \\ (6,5,3) \end{array}\right. \\ & \Rightarrow 8 \times 3!+2(3!-2!)=48+8=56 \\ & =0+14+56=70 \end{aligned}$
The number of 3-digit numbers, formed using the digits 2, 3, 4, 5 and 7, when the repetition of digits is not allowed, and which are not divisible by 3 , is equal to ________.
Explanation:
To solve this problem, we need to find the number of 3-digit numbers formed using the digits 2, 3, 4, 5, and 7, with no repetition of digits allowed, and these numbers should not be divisible by 3. Let's break down the solution step-by-step:
1. Calculating the total number of 3-digit numbers without repetition:
The number of ways to form a 3-digit number from 5 unique digits (2, 3, 4, 5, 7) without repetition can be calculated using permutations:
The total number of permutations for choosing 3 digits out of 5 and arranging them is given by:
$5P3 = \frac{5!}{(5-3)!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} = 5 \times 4 \times 3 = 60$
So, there are 60 possible 3-digit numbers that can be formed from the digits 2, 3, 4, 5, and 7 without repetition.
2. Finding the 3-digit numbers divisible by 3:
A number is divisible by 3 if the sum of its digits is divisible by 3. Let's consider the sums of every combination of these three digits to find out which sums are divisible by 3:
Possible sums of combinations:
- 2 + 3 + 4 = 9 (divisible by 3)
- 2 + 4 + 7 = 13
- 2 + 5 + 7 = 14
- 2 + 3 + 5 = 10
- 2 + 3 + 7 = 12 (divisible by 3)
- 3 + 4 + 5 = 12 (divisible by 3)
- 3 + 4 + 7 = 14
- 3 + 5 + 7 = 15 (divisible by 3)
- 4 + 5 + 7 = 16
The combinations whose sums are divisible by 3 are:
- 2, 3, 4
- 2, 3, 7
- 3, 4, 5
- 3, 5, 7
Since the sum of the digits is divisible by 3 for these combinations, any permutation of these sets will yield a number divisible by 3:
The number of 3-digit numbers that can be formed from each set of 3 digits is:
$3! = 6$
So, the total number of 3-digit numbers divisible by 3 is:
$4 \text{ sets} \times 6 \text{ permutations per set} = 24$
3. Calculating the 3-digit numbers not divisible by 3:
To find the 3-digit numbers not divisible by 3, we subtract the number of those divisible by 3 from the total number of 3-digit numbers:
$60 - 24 = 36$
Therefore, the number of 3-digit numbers that can be formed using the digits 2, 3, 4, 5, and 7 without repetition, and which are not divisible by 3, is equal to 36.
The number of ways of getting a sum 16 on throwing a dice four times is ________.
Explanation:
Number of ways $=$ coefficient of $x^{16}$ in $\left(x+x^2+\ldots+\right.$ $\left.x^6\right)^4$
$=$ coefficient of $x^{16}$ in $\left(1-x^6\right)^4(1-x)^{-4}$
$=$ coefficient of $x^{16}$ in $\left(1-4 x^6+6 x^{12} \ldots\right)(1-x)^{-4}$
$={ }^{15} C_3-4 \cdot{ }^9 C_3+6=125$
There are 4 men and 5 women in Group A, and 5 men and 4 women in Group B. If 4 persons are selected from each group, then the number of ways of selecting 4 men and 4 women is ________.
Explanation:
| Group A | Group B | Ways |
|---|---|---|
| $4m$ $3m+1w$ $2m+2w$ $1m+3w$ $4w$ |
$4w$ $1m+3w$ $2m+2w$ $3m+w$ $4m$ |
${ }^4 C_4 \cdot{ }^4 C_4 \quad=1$ ${ }^4 C_1 \cdot{ }^5 C_1 \cdot{ }^5 C_1^4 C_3 \quad=400$ ${ }^4 C_2 \cdot{ }^5 C_2{ }^5 C_2{ }^4 C_2 \quad=3600$ ${ }^4 C_1{ }^5 C_3{ }^5 C_3{ }^4 C_1 \quad=1600$ ${ }^5 C_4{ }^5 C_4 \quad=25$ |
$\begin{aligned} \text { Total ways } & =1+400+3600+1600+25 \\ & =5626 \end{aligned}$
Explanation:
$ x, y, z \geq 0 $
as
$\begin{array}{ll}z=0 & x+2 y=42 \Rightarrow 22 \text { cases } \\\\ z=1 & x+2 y=39 \Rightarrow 20 \text { cases } \\\\ z=2 & x+2 y=36 \Rightarrow 19 \text { cases } \\\\ z=3 & x+2 y=33 \Rightarrow 17 \text { cases } \\\\ z=4 & x+2 y=30 \Rightarrow 16 \text { cases } \\\\ z=5 & x+2 y=27 \Rightarrow 14 \text { cases }\end{array}$
$z=6 \quad x+2 y=24 \Rightarrow 13$ cases
$z=7 \quad x+2 y=21 \Rightarrow 11$ cases
$z=8 \quad x+2 y=18 \Rightarrow 10$ cases
$z=9 \quad x+2 y=15 \Rightarrow 8$ cases
$z=10 x+2 y=12 \Rightarrow 7$ cases
$z=11 x+2 y=9 \Rightarrow 5$ cases
$z=12 x+2 y=6 \Rightarrow 4$ cases
$z=13 x+2 y=3 \Rightarrow 2$ cases
$z=14 x+2 y=0 \Rightarrow 1$ case
Total = 169
The total number of words (with or without meaning) that can be formed out of the letters of the word 'DISTRIBUTION' taken four at a time, is equal to __________.
Explanation:
We have III, TT, D, S, R, B, U, O, N
Number of words with selection (a, a, a, b)
$={ }^8 \mathrm{C}_1 \times \frac{4 !}{3 !}=32$
Number of words with selection (a, a, b, b)
$=\frac{4 !}{2 ! 2 !}=6$
Number of words with selection (a, a, b, c)
$={ }^2 \mathrm{C}_1 \times{ }^8 \mathrm{C}_2 \times \frac{4 !}{2 !}=672$
Number of words with selection (a, b, c, d)
$\begin{aligned} & ={ }^9 \mathrm{C}_4 \times 4 !=3024 \\\\ & \therefore \text {Total }=3024+672+6+32 \\\\ & =3734 \end{aligned}$
In an examination of Mathematics paper, there are 20 questions of equal marks and the question paper is divided into three sections : $A, B$ and $C$. A student is required to attempt total 15 questions taking at least 4 questions from each section. If section $A$ has 8 questions, section $B$ has 6 questions and section $C$ has 6 questions, then the total number of ways a student can select 15 questions is __________.
Explanation:
The problem involves choosing 15 questions out of a total of 20 available questions, with the constraint that at least 4 questions must be chosen from each of the three sections A, B, and C. To evaluate the total number of ways a student can select these questions, we need to consider every possible combination of questions from sections A, B, and C that sum up to 15 questions while respecting the constraints.
Section A has 8 questions, Section B and C each have 6 questions. The student must choose at least 4 questions from each section, which satisfies the minimum requirement. However, since the student is to attempt a total of 15 questions, there are several combinations to consider, as outlined below:
- Choosing 4 questions from A, 5 from B, and 6 from C
- Choosing 4 questions from A, 6 from B, and 5 from C
- Choosing 7 questions from A, 4 from B, and 4 from C
- Choosing 6 questions from A, 5 from B, and 4 from C
- Choosing 6 questions from A, 4 from B, and 5 from C
- Choosing 5 questions from A, 5 from B, and 5 from C
- Choosing 5 questions from A, 6 from B, and 4 from C
- Choosing 5 questions from A, 4 from B, and 6 from C
$ \begin{array}{|c|c|c|l|c|} \hline \mathrm{A} & \mathrm{B} & \mathrm{C} & \Rightarrow & \begin{array}{c} \text { No. of } \\ \text { question } \end{array} \\ \hline 4 & 5 & 6 & \rightarrow & { }^8 C_4{ }^6 C_5{ }^6 C_6 \\ \hline 4 & 6 & 5 & \rightarrow & { }^8 C_4{ }^6 C_6{ }^6 C_5 \\ \hline 7 & 4 & 4 & \rightarrow & { }^8 C_7{ }^6 C_4{ }^6 C_4 \\ \hline 6 & 5 & 4 & \rightarrow & { }^8 C_6{ }^6 C_5{ }^6 C_4 \\ \hline 6 & 4 & 5 & \rightarrow & { }^8 C_6{ }^6 C_4{ }^6 C_5 \\ \hline 5 & 5 & 5 & \rightarrow & { }^8 C_5{ }^6 C_5{ }^6 C_5 \\ \hline 5 & 6 & 4 & \rightarrow & { }^8 C_5{ }^6 C_6{ }^6 C_4 \\ \hline 5 & 4 & 6 & \rightarrow & { }^8 C_5{ }^6 C_4{ }^6 C_6 \\ \hline \end{array} $
Total ways of select $=11376$
All the letters of the word "GTWENTY" are written in all possible ways with or without meaning and these words are written as in a dictionary. The serial number of the word "GTWENTY" is _________.
Explanation:
Words starting with $\mathrm{E}=360$
Words starting with $\mathrm{GE=60}$
Words starting with $\mathrm{GN=60}$
Words starting with $\mathrm{GTE=24}$
Words starting with $\mathrm{GTN=24}$
Words starting with $\mathrm{GTT=24}$
GTWENTY $=1$
Total $=553$
The number of ways five alphabets can be chosen from the alphabets of the word MATHEMATICS, where the chosen alphabets are not necessarily distinct, is equal to:
Let $[t]$ be the greatest integer less than or equal to $t$. Let $A$ be the set of all prime factors of 2310 and $f: A \rightarrow \mathbb{Z}$ be the function $f(x)=\left[\log _2\left(x^2+\left[\frac{x^3}{5}\right]\right)\right]$. The number of one-to-one functions from $A$ to the range of $f$ is
If all the words with or without meaning made using all the letters of the word "NAGPUR" are arranged as in a dictionary, then the word at $315^{\text {th }}$ position in this arrangement is :
Let $0 \leq r \leq n$. If ${ }^{n+1} C_{r+1}:{ }^n C_r:{ }^{n-1} C_{r-1}=55: 35: 21$, then $2 n+5 r$ is equal to :
The number of triangles whose vertices are at the vertices of a regular octagon but none of whose sides is a side of the octagon is
Let the set $S=\{2,4,8,16, \ldots, 512\}$ be partitioned into 3 sets $A, B, C$ with equal number of elements such that $\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}=\mathrm{S}$ and $\mathrm{A} \cap \mathrm{B}=\mathrm{B} \cap \mathrm{C}=\mathrm{A} \cap \mathrm{C}=\phi$. The maximum number of such possible partitions of $S$ is equal to:
60 words can be made using all the letters of the word $\mathrm{BHBJO}$, with or without meaning. If these words are written as in a dictionary, then the $50^{\text {th }}$ word is:
There are 5 points $P_1, P_2, P_3, P_4, P_5$ on the side $A B$, excluding $A$ and $B$, of a triangle $A B C$. Similarly there are 6 points $\mathrm{P}_6, \mathrm{P}_7, \ldots, \mathrm{P}_{11}$ on the side $\mathrm{BC}$ and 7 points $\mathrm{P}_{12}, \mathrm{P}_{13}, \ldots, \mathrm{P}_{18}$ on the side $\mathrm{CA}$ of the triangle. The number of triangles, that can be formed using the points $\mathrm{P}_1, \mathrm{P}_2, \ldots, \mathrm{P}_{18}$ as vertices, is:
