Permutation and combination questions

Given alphabets are $A \ , \ H \ , \ L \ , R \ , \ U$
Number of words beginning with $A = 4! = 24$
Number of words beginning with $H = 4! = 24$
Number of words beginning with $L = 4! = 24$
first word beginning with $R$ is $RAHLU$
Next word is $RAHUL$

$\therefore$ its Rank $= 24+24+24+1+1$
$=74$

The $7$ men can be seated at the round table in $6!$ ways.

In each of these seating's ,the $5$ ladies can be seated in between the men in $^7P_5$ ways. This can together be done in $6! \times ^7P_5$ ways.

Number of ways of seating $= 6! \times ^7P_5$

$=\frac{7}{2} (6!)^2=\frac{7}{2}(720)^2$

We know $C_0^2+C_1^2+ \dots + C_n^2= \large \frac{(2n)!}{(n!)^2}$

put $n=5.$ Then $C_0^2+C_1^2+ \dots + C_5^2= \large \frac{10!}{(5!)^2}$ $=\large \frac{10!}{5! \times 5!} = \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5!}{5 \times 4 \times 3 \times 2 \times 1 \times 5!}=252$

$\frac{50C_0}{1}+\frac{50C_2}{3}+\frac{50C_4}{5}+ \dots + \frac{50C_{50}}{51}$

$= \frac{1}{51} \left [\frac{51}{1}+\frac{51 \times 50 \times 49}{1 \times 2 \times 3} +\frac{51 \times 50 \times 49 \times 48 \times 47 }{1 \times 2 \times 3 \times 4 \times 5}+ \dots \right]$

$= \frac{1}{51}[ \ ^{51}C_1+ \ ^{51}C_3+ \ ^{51}C_5+ \dots+ \ ^{51}C_{51}]$

$= \large \frac{1}{51} \times \frac{2^{51}}{2}=\frac{2^{50}}{51}$

The sum is even if either
$\hspace{6mm}i)$ all the $3$ numbers are even.
or $ii)$ one is even and two are odd.

$\therefore$ No. of ways of selecting the numbers

$=C(15,3)+C(15,1) \times C(15,2)$

$=455+1575$
$=2030$

Given $^{18}C_{15}+2. \ ^{18}C_{16}+ \ ^{17}C_{16+1}= \ ^nC_3$

ie$, \ ^{18}C_{3}+2.( \ ^{18}C_{2})+ \ ^{17}C_{1}+1= \ ^nC_3$

ie$, \ ^{18}C_{16}+306+17+1= \ ^nC_3 \left [ \ ^nC_r = ^nC_{n-r} \right]$

$\therefore ^nC_3=1140$

$\therefore n = 20$

A regular polygon of $n+3$ sides have $n+3$ vertices
$\therefore$ Number of triangles that can be formed $= c(n+3,3)$
$\therefore$ Given $c(n+3,3)=220$

i.e,$\ \large \frac{(n+3)(n+2)(n+1)}{1 \times 2 \times 3}=220$

i.e,$\ (n+3)(n+2)(n+1) = \\ 220 \times 6 \left[ \because 220 \times 6 = 22 \times 10 \times 6=2 \times 11 \times 10 \times 6 \right]$

$= 10 \times 11 \times 12$

$\therefore n+1=10$

$\therefore n=9$

$\frac{1}{ ^4C_r}=\frac{1}{ ^5C_r}+\frac{1}{ ^6C_r}$

$\Rightarrow \frac{(4-r)!r!}{ 4!}= \frac{(5-r)!r!}{ 5!}+ \frac{(6-r)!r!}{ 6!}$ $\left[\because ^nC_r= \frac{n!}{(n-r)!r!}\right]$

$\Rightarrow \frac{(4-r)!}{ 4!}=\frac{(5-r)!}{ 5!}+\frac{(6-r)!}{ 6!}$

$\Rightarrow \frac{(4-r)!}{ 4!}=\frac{(5-r)(4-r)!}{5 \times 4!}+\frac{(6-r)(5-r)(4-r)!}{6 \times 5 \times 4!}$

$\Rightarrow 1=\frac{5-r}{5}+\frac{(6-r)(5-r)}{30}$

$\Rightarrow 30-6r+30-6r-5r+r^2=30$

$\Rightarrow 30-6r+30-11r+r^2=30$

$\Rightarrow r^2-17r+30=0$

$\Rightarrow r^2-15r-2r+30=0$

$\Rightarrow r(r-15)-2(r-15)$

$\Rightarrow (r-15)(r-2)$

$\Rightarrow r=2 \ , \ \because( r\neq 15)$ (r cannot be greater than 4)

$\ ^{12}C_3-\left[ \ ^3C_3+ \ ^4C_3+ \ ^5C_3 \right]$

$\Rightarrow \frac{12!}{3! \ 9!}- \left[ \frac{3!}{3!}+\frac{4!}{3!}+\frac{5!}{3! \ 2!} \right]$

$\Rightarrow \frac{ 12 \times 11 \times 10 \times 9!}{3 \times 2 \times 9!}- \left[1+\frac{4 \times 3!}{3!}+\frac{5 \times 4 \times 3!}{2 \times 3!} \right]$

$\Rightarrow 220-[1+4+10]$

$\Rightarrow 220-15$

$\Rightarrow 205$

$8! \left [ \frac{1}{3!}+ \frac{5}{4!} \right]=\frac{8!}{3!}\left [1+\frac{5}{4}\right ]$

$=\frac{8!}{3!} \times \frac{9}{4}=\frac{9!}{24}$

$\frac{9!}{24}= \ ^9P_r = (9-r)!=4!$

$\Rightarrow 9-r=4$

$\Rightarrow r=5$

$7 \times 6 \times 5 =210$

Numbers can be with $1$ digit, $2$ digit and $3$ digits are $1,2,3,4,5,6,7$

Required numbers $= \ ^7P_1+ \ ^7P_2+ \ ^7P_3$

$\large \frac{7!}{6!}+\frac{7!}{5!}+\frac{7!}{4!}$

$\large \frac{7 \times 6!}{6!}+\frac{7 \times 6 \times 5!}{6!}+\frac{7 \times 6 \times 5 \times 4!}{6!}$

$=259$

Given $P(n,r) = 1680$ and $C(n,r)=70$

$\therefore \frac{P(n,r)}{C(n,r)}=\frac {1680}{70}$

$\therefore r!=24$ $\therefore r=4$

$P(n,r) = 1680 \Rightarrow P(n,4) = 1680$

$\Rightarrow n(n-1)(n-2)(n-3)=5 \times 6 \times 7 \times 8$

$\Rightarrow n= 8$

$\therefore 69n+r! = 69 \times 8+24=576$

If the middle digit is $K(2 \leq K \leq 9)$, then the $100^{th}$ digit can be chosen in$K-1$ ways ( as this digit cannot be zero ) and unit's digit can be chosen in $K$ ways .
Thus number of required numbers with middle digit is bigger than the extreme digit is $(K-1)K$
As $2 \leq K \leq 9$
$\therefore$ Number of required numbers is $\sum_{K=2}^9 K(K-1)=240$

Number of words begining with $E= 5! = 120$
Number of words begining with $H= 5! = 120$
Number of words begining with $ME= 4! = 24$
Number of words begining with $MH= 4! = 24$
Number of words begining with $MOE= 3! = 6$
Number of words begining with $MOH= 3! = 6$
Number of words begining with $MOR= 3! = 6$
Number of words begining with $MOTE= 2! = 2$
Next word is MOTHER
$\therefore$ its rank $=120+120+24+24+6+6+6+2+1=309$

$^nP_r=3024$ and $^C_r=126$

$\large \frac{n!}{(n-r)!}$$=3024$ and $\large\frac{n!}{r! \ (n-r)!}$$=126$ $\Rightarrow r!=24$

$\Rightarrow r!=4!$

$\Rightarrow r=4$

A committee pf 4 persons can be formed in following ways:

(i) 3 persons and one ladies = $^7 C_3$$\times$$^3 C_1$ =35×3=105

(ii) 2 persons and two ladies = $^7 C_2$ × $^3 C_2$ =21×3=63

(iii) 1 persons and three ladies = $^7 C_1$ × $^3 C_3$ =7×1=7

Therefore, total number of ways =105+63+7=175

Given $^nC_3=220$

$\Rightarrow \large \frac{n(n-1)(n-2)(n-3)!}{3! \ (n-3)!} =220$

$\Rightarrow \large \frac{n(n-1)(n-2)}{3 \times 2\times 1}=220$

$\Rightarrow n(n-1)(n-2)=1320$
$\Rightarrow n(n-1)(n-2) = 12.11.10$
$\Rightarrow n=12$

Let the total no $:$ of stones be $2n+1$ (ie) there are $n$ stone to the left of middle stone and $n$ to its right.
Since the man starts one point$,$ total distance traveled by him for collecting all stone to the middle stone is
$4 \left [10 \times n +10(n-1)+ \dots + 10 \times 1 \right ] - 10 \times n= 3000$ $[$As $3 \ km=3000 \ m]$

$\Rightarrow 10 \left [ 4 \ [ n +(n-1)+ \dots + 1] \ - n \right ]= 3000$

$\Rightarrow [ 4 \times \frac{n}{2} (n+1) - n] = 300$

$\Rightarrow 2n^2+2n-n=300$
$\Rightarrow 2n^2+n-300=0$
$\Rightarrow 2n^2+25n-24n-300=0$
$\Rightarrow n(2n+25)-12(2n+25)=0$
$\Rightarrow (n-12) \ (2n+25) = 0$
$\Rightarrow n = 12 \ , \ n= \frac {-25}{2}$
Now$, \ n \neq \frac{-25}{2}, \ \therefore n= 12$
Hence total no. of stone $=2n+1=24+1=25$

$\frac{729+6 \times 2 \times 243 + 15 \times 4 \times 81 + 20 \times 8 \times 27 \\ +15\times 16\times 9 + 6 \times 32 \times 3 + 64}{1+ 4\times 4 + 6 \times 16 + 4 \times 64 + 256}$=$\frac{(3+2)^6}{(1+4)^4} \Rightarrow x=25$
$\sqrt {25} - \frac 1{\sqrt {25}}=4.8$

$(n+2)!=2550 \ n!$
$\Rightarrow (n+2) \ (n+1) \ n! = 2550 \ n!$
$\Rightarrow n^2+2n+n+2=2550$
$\Rightarrow n^2+3n-2548=0$
$\Rightarrow n^2+52n-49n-2548=0$
$\Rightarrow n(n+52)-49(n+52)$
$\Rightarrow (n+52) \ (n-49)$
$\Rightarrow n=49$

There can be two types of committees.
$i)$ Containing $3$ men and $3$ ladies $= \ ^8C_3 \times ^4C_3$

$= \large \frac{8!}{3! \ 5!} \times \frac{4!}{3!}$

$= \large \frac{8 \times 7 \times 6 \times 5!}{3 \times 2 \times 5!} \times \frac{4 \times 3!}{3!}$

$= 224$

$i)$ Containing $2$ men and $4$ ladies $= \ ^8C_2 \times ^4C_4$

$= \large \frac{8!}{2! \ 6!} \times \frac{4!}{4!}$

$= \large \frac{8 \times 7 \times 6!}{2 \times 6! } \times 1$

$= 28$

$\therefore$ required no $:$ of ways $=224+28=252$

Requierd no $:$ of straight lines $= \ ^{20}C_2 - \ ^4C_2 +1$
$= \frac{20!}{2! \ 18!} - \frac{4!}{2! \ 2!}+1$

$=\large \frac{20 \times 19 \times 18!}{2 \times 18!} - \frac{4 \times 3 \times 2!}{2 \times 2!} +1$

$\Rightarrow \large \frac{20 \times 19}{2}- \frac{4 \times 3}{2}+1$

$\Rightarrow 190-6+1$
$=185$

Since $^{43}C_{r-6} = \ ^{43}C_{3r+1}$

either $r-6=3r+1$ or $r-6+3r+1=43$

$\Rightarrow$ either $2r=-7$ or $4r=48$

$\Rightarrow$ either $r= \frac{-7}{2}$ or $r=12$

But $r =\frac{-7}{2}$ is not possible$, \ \therefore r = 12$

$\large \frac{ \ ^{56}P_r+6}{ \ ^{54}P_r+3} =\frac{30800}{1}$

$\Rightarrow \large \frac{56!}{(50-r)!} \times \frac{(51-r)!}{54!}=30800$

$\Rightarrow \large \frac{56 \times 55 \times 54! (51-r) \ (50-r)!}{(50-r)! \ 54!}= 30800$

$\Rightarrow 56 \times 55 \ (51-r) =30800$

$\Rightarrow 51-r=10$

$\Rightarrow r=41$

First we seat $5$ boys. This can be done in $5!$ ways$, ie$
this can be done in $120$ ways.
$B_1.B_2.B_3.B_4.B_5$
The girls can be seated on the four places marked $'.',$
so that each girl is between two boys.
This can be done in $^{4}P_3$ ways $= 4 \times 3 \times 2 = 24$
Hence required number of ways $= 120 \times 24 = 2880$

$\ ^{n+1}C_3 - \ ^nC_3 = 28$

$\Rightarrow \large \frac{n(n-1)(3)}{6}=28$

$\Rightarrow 3n^2-3n-168=0$

$\Rightarrow n^2-n-56=0$

$\Rightarrow n^2 -8n+7n-56=0$

$\Rightarrow n(n-8)+7(n-8)=0$

$\Rightarrow (n-8)(n+7)=0$

$\Rightarrow n=8 \ , \ n=-7$

$\Rightarrow n=8$

fix $0$ at the units place
no $:$ of four digit even numbers $= \ ^{6}P_3 = \large \frac{6!}{3!}=\frac{6 \times 5 \times 4 \times 3!}{3!}$
$= 6 \times 5 \times 4 = 120$

fix $2$ or $4$ or $6$ at the units place
no $:$ of four digit even numbers $= 3 \left[ \ ^{6}P_3 - \ ^{5}P_2 \right]$

$=\large 3 \left [ \frac{6!}{3!} - \frac{5!}{2!} \right] = 3 \left [ \frac{6 \times 5 \times 4 \times 3!}{3!} - \frac{5 \times 4 \times 3 \times 2!}{2!} \right]$

$= 3[120-20]= 300$

$\therefore$ total number of ways $= 300+120 = 420$

Total number of hand shakes $= \ ^{n}C_2 = 66$
$n$ is the number of persons.

$\Rightarrow\large \frac{n(n-1)}{2} =66$

$\Rightarrow n^2-n-132=0$

$\Rightarrow n^2-12n+11n-132=0$

$\Rightarrow n(n-12)+11(n-12)=0$

$\Rightarrow (n-12)(n+11)$

$\Rightarrow n=12 \ , \ n= -11$

$\Rightarrow n=12$

Odd numbers are 1, 3, 5, 7
We have to fill up four places like TH H T U ( if repetition is allowed )
$^ 5 C_1 \space 6^2 \space ^4 C_1 = 5 \times 6^2 \times 4= 5 \times 36 \times 4=720$

Let the number of digits formed be $x$
1000 < x < 4000, which means left extreme digit will be either 2 or 3
Required numbers = $^2 C_1 \times H \space T \space U$
Where, H = Hundred's place, T = Ten's place and U =Unit's place
= $^2 C_1 \times 4 \times 4 \times 4 =128$

Consider $^n C_{r-1} + 2 \space ^n C_r + ^n C_{r+1}$
= $(^n C_{r-1} + ^n C_r) + ( ^n C_r + ^n C_{r+1})$
= $^{n+1} C_r + ^{n+1} C_{r+1} = ^{n+2} C_{r+1}$

Number of women =5
Number of Men =6
Number of ways of 6 men at a round table is $(n-1)!= (6-1)! = 5!$
Now we left with six places between the men and there are 5 women, these 5 women can be arranged themselves by $^6 P_5$ ways
Required number of ways = $5! \times ^6 P_5= 5! \times 6!$

No of letters = 6
Number of vowels = 2 , namely A & E these alphabets can be arrange themselves by 2! ways
Number of words = $\frac{6!}{2!}=360$

$^{50} C_4 + \sum_{r=1} ^6 \space ^{56-r} C_3$
putting $r=6, 5, 4, 3, 2, 1$, we get
$^{50} C_4 + ^{50} C_3 + ^{51} C_3 + ^{52} C_3 + ^{53} C_3+ ^{54} C_3 + ^{55} C_3$
$( ^n C_r + ^n C_{r+1} = ^{n+1} C_{r+1})$
$=^{51} C_4 + ^{51} C_3 + ^{52} C_3 + ^{53} C_3 + ^{54} C_3+ ^{55} C_3$
$=^{52} C_4 + ^{52} C_3 + ^{53} C_3 + ^{54} C_3 + ^{55} C_3$
$^{53} C_4 + ^{53} C_3 + ^{54} C_3 + ^{55} C_3 = ^{54} C_4+ ^{54} C_3 + ^{55} C_3$
= $^{55} C_4 + ^{55} C_3 = ^{56} C_4$

'5' balls to be distributed among '3'. Total number of distribution possible =3×3×.....5times =$3^5)$.

Of these distributions, there are cases when one person does not get any ball. Number of such cases = $^{3}C_{1}$$\times$ $2^{5}-2$ =90

Also, there are cases when one gets all. Number of such cases= $^{3}C_{1}$=3

So, distributions such that each gets at least 1 is

$3^{5}$ −90−3 =243−93 =150.

In the word $SACHIN$, $S$ is fixed
No of words starts with A = 5!
No of words starts with C = 5!
No of words starts with H= 5!
No of words starts with I = 5!
No of words starts with N = 5!
Total words = 5!+ 5!+ 5!+5!+5!= 5(5!)=600
Now add the rank of $SACHIN$ so required rank of $SACHIN$ = $600+1=601$

$^{20} C_0 + ^{20} C_1 x + ......+ ^{20} C_{10} x^{10}+ ........ + ^{20} C_{20} x^{20}= (1+x)^{20}$
After putting $x=-1$ , we get $^{20} C_0 - ^{20} C_1 + ^{20} C_2 - ^{20} C_3 + .......+ ^{20} C_{10} - ^{20} C_{11} - ^{20} C_{12} + .....^{20} C_{20}=0$
$2(^{20} C_0 - ^{20} C_1 + ^{20} C_2 - ^{20} C_3 + .....- ^{20} C_9)+ ^{20} C_{10}=0$
$^{20} C_0 - ^{20} C_1 + ^{20} C_2 - ^{20} C_3 + ...... - ^{20} C_9 + ^{20} C_{10}= \frac{1}{2} \space ^{20} C_{10}$

Leaving $S$, we have 7 letters M, I, I, I, P, P, I
ways of arranging them = $\frac{7!}{2!.4!}= 7.5.3$ And four $S$ can be put in 8 places in $^8C_4$ ways. The required number of ways = $7.5.3. ^8 C_4= 7. ^6 C_4. ^8C_4$

We have to find the number of integral solutions if $x_1 + x_2 + x_3 +x_4 +x_5 =6$ and that equals $^{5+6-1} C_{5-1} = ^{10} C_4$
Thus statement 1 is false
Number of different ways of arranging 6 A's and 4 B's in a row = $\frac{10!}{6! \times 4!}= ^{10} C_4$= Number of different ways the child can buy the six ice creams .
Statement 2 is true
so, statement 1 is false ; statement 2 is true

Out of 6 novels , 4 novels can be selected in $^6 C_4$ ways. Also out of 3 dictionaries 1 dictionary can be selected in $^3 C_1$ ways.
Since the dictionary is fixed in the middle we only have to arrange 4 novels which can be done in $4!$ ways
Then the number of ways = $^6 C_4. ^3 C_1. 4!$
= $\frac{6.5}{2} . 3. 24 =1080$

• Cis-2-butene and trans-2-butene are stereoisomers because they are different compounds with different spatial arrangements of their atoms:
• Cis-2-butene: The methyl groups are on the same side of the double bond.
• Trans-2-butene: The methyl groups are on opposite sides of the double bond. .
• Cis-2-butene and trans-2-butene are also known as geometric isomers. They have different physical, chemical, and physiological properties.

$S_1 = \sum j(j-1) \space ^{10} C_j$
$= \sum j(j-1) . \frac{10.(10-1)}{j(j-1)}. \space ^8 C_{j-2}$
= $9 \times 10 \sum _{j=2} ^{10} \space ^8C_{j-2} = 90 \times 2^8$

$S_2 = \sum _{j=1} ^{10} \space j . \space ^{10} C_j = 10 \sum _{j=1} ^{10} \space ^9 C_{j=1} 10 \times 2^9$
$S_3 = \sum_{j=1} ^{10} \space j^2. \space ^{10} C_j = \sum_{j=1} ^{10} \space(j (j-1)+j). ^{10} C_j$
$= \sum_{j=1} ^{10} j(j-1) \space ^{10} C_j + \sum_{j=1} ^{10} j. \space ^{10} C_j$
$= 90.2^8 + 10. 2^9 = (45+10)2^9 =55.2^9$
Thus statement 1 is true and statement 2 is false

We can have either 4 digit or 5 digit number.

First consider 4 digit number.

We can take 6,7,8 at the thousandth place.

The next three places can be filled in $^{4}C_{3}$ ⋅3! way.

So we get 72 such numbers

For 5 digit number, we can choose them in 5! ways(120).

So total no of such numbers are 72+120=192.

$x_1+x_2+x_3+x_4=10$
The number of positive integral solution is $^{6+4-1} C_{4-1}= ^9 C_3$
It is the same as the number of ways of choosing any 3 places from 9 different places

Number of ways of selecting 3 elements in $A$ such that $f(x)=y_2$ is $^7 C_3$
Now for remaining 4 elements in $A$, we have 2 elements in $B$
Total number of onto functions
= $^7 C_3 \times (2^4- ^2 C_1(2-1)^4)= ^7 C_3 \times 14$