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 = \( ^6 C_4. ^3 C_1. 4!\)
= \( \frac{6.5}{2} . 3. 24 =1080\)

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

\(S_2 = \sum _{j=1} ^{10} \space j . \space ^{10} C_j = 10 \sum _{j=1} ^{10} \space ^9 C_{j=1} 10 \times 2^9\)
\(S_3 = \sum_{j=1} ^{10} \space j^2. \space ^{10} C_j = \sum_{j=1} ^{10} \space(j (j-1)+j). ^{10} C_j \)
\( = \sum_{j=1} ^{10} j(j-1) \space ^{10} C_j + \sum_{j=1} ^{10} j. \space ^{10} C_j \)
\( = 90.2^8 + 10. 2^9 = (45+10)2^9 =55.2^9\)
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
= \(^7 C_3 \times (2^4- ^2 C_1(2-1)^4)= ^7 C_3 \times 14\)