Permutations and Combinations

Case 1 :

Among four bags from one bag 4 balls are taken. Number of ways of chosing one bag from 4 bags = ${}^4{C_1}$

From this bag number of ways of taking 4 balls are

(a) 3 Red and 1 Blue balls, which can be chosen in ${}^3{C_3} \times {}^2{C_1}$ ways.

(b) 2 Red and 2 Blue balls, which can be chosen in ${}^3{C_2} \times {}^2{C_2}$ ways.

$ \therefore $ Total number of ways of choosing this 4 balls from this bag = ${}^4{C_1}$$\left( {{}^3{C_3} \times {}^2{C_1} + {}^3{C_2} \times {}^2{C_2}} \right)$

Now, two balls are taken from remaining three bags.

From each bag two balls can be taken in ${{}^3{C_1} \times {}^2{C_1}}$ ways.

So, for three bags two balls can be taken in ${\left( {{}^3{C_1} \times {}^2{C_1}} \right)^3}$ ways.

$ \therefore $ Total number of ways of chosing 10 balls from these four bags

= ${}^4{C_1}$$\left( {{}^3{C_3} \times {}^2{C_1} + {}^3{C_2} \times {}^2{C_2}} \right)$ $ \times $ ${\left( {{}^3{C_1} \times {}^2{C_1}} \right)^3}$

Case 2 :

Among four bags from two bags 3 balls are taken. Number of ways of chosing two bags from 4 bags = ${}^4{C_2}$

From each bag, number of ways of taking 3 balls are

(a) 2 Red and 1 Blue balls, which can be chosen in ${}^3{C_2} \times {}^2{C_1}$ ways.

(b) 1 Red and 2 Blue balls, which can be chosen in ${}^3{C_1} \times {}^2{C_2}$ ways.

So, for two bags three balls can be taken in ${\left( {{}^3{C_2} \times {}^2{C_1} + {}^3{C_1} \times {}^2{C_2}} \right)^2}$ ways.

$ \therefore $ Total number of ways of choosing this 3 balls from this two bags = ${}^4{C_2}$${\left( {{}^3{C_2} \times {}^2{C_1} + {}^3{C_1} \times {}^2{C_2}} \right)^2}$

Now, two balls are taken from remaining two bags.

From each bag two balls can be taken in ${{}^3{C_1} \times {}^2{C_1}}$ ways.

So, for two bags two balls can be taken in ${\left( {{}^3{C_1} \times {}^2{C_1}} \right)^2}$ ways.

$ \therefore $ Total number of ways of chosing 10 balls from these four bags

= ${}^4{C_2}$${\left( {{}^3{C_2} \times {}^2{C_1} + {}^3{C_1} \times {}^2{C_2}} \right)^2}$ $ \times $ ${\left( {{}^3{C_1} \times {}^2{C_1}} \right)^2}$

From Case 1 and Case 2, total number of ways of chosing 10 balls from these 4 boxes so that from each box at least one red ball and one blue ball are chosen

= ${}^4{C_1}$$\left( {{}^3{C_3} \times {}^2{C_1} + {}^3{C_2} \times {}^2{C_2}} \right)$ $ \times $ ${\left( {{}^3{C_1} \times {}^2{C_1}} \right)^3}$

+ ${}^4{C_2}$${\left( {{}^3{C_2} \times {}^2{C_1} + {}^3{C_1} \times {}^2{C_2}} \right)^2}$ $ \times $ ${\left( {{}^3{C_1} \times {}^2{C_1}} \right)^2}$

$ \begin{aligned} & =4(5)(6)^3+6(3 \times 2+3)^2(6)^2 \\\\ & =4320+17496 \\\\ & =21816 \end{aligned} $

Given, ${ }^m P_r-{ }^{(m-1)} P_r=a \cdot{ }^{(m-1)} P_s$

$ \Rightarrow{ }^m P_r={ }^{(m-1)} P_r+a^{(m-1)} P_s \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Also, we know that

$ { }^n P_r={ }^{n-1} P_r+r \cdot{ }^{n-1} P_{r-1} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

On comparing Eqs. (i) and (ii), we get

$ \begin{aligned} & a=r \\ \text { and } s=r-1 & \\ \Rightarrow \quad & a-s=r-(r-1) \\ \Rightarrow \quad & a-s=1 \end{aligned} $

Given, word is TSEAMCET

In the word, T and E occur twice. So, we have S, A, M, C, TT, EE. We have 6 distinct letter.

Way 14 letters selected and all are distinct.

$ { }^6 C_4={ }^6 C_2=\frac{6 \times 5}{2 \times 1}=15 $

Way 22 letters are alike and 2 are distinct

$ { }^2 C_1 \times{ }^5 C_2=2 \times \frac{5 \times 4}{2 \times 1}=20 $

Way 32 letters are alike and rest 2 are also alike $\quad{ }^2 C_2=1$

So, the total number of ways

$ =15+20+1=36 $

Given, $a+b+c=5$ and $a, b, c \in N L$ is least value of $2^a 3^b 5^c$ and $M$ is the greatest value of $2^a 3^b 5^5$.

Possible pairs of $(a, b, c)$

$ \begin{array}{r} =(1,1,3),(1,3,1),(3,1,1),(1,2,2) \\ (2,1,2),(2,2,1) \end{array} $

Value for $(1,1,3)=\left(2 \cdot 3 \cdot 5^3\right)=750$

Value for $(1,3,1)=\left(2 \cdot 3^3 \cdot 5\right)=270$

Value for $(3,1,1)=\left(2^3 \cdot 3 \cdot 5\right)=120$

Value for $(1,2,2)=\left(2^1 \cdot 3^2 \cdot 5^2\right)=450$

Value for $(2,1,2)=\left(2^2 \cdot 3^1 \cdot 5^2\right)=300$

Value for $(2,2,1)=\left(2^2 \cdot 3^2 \cdot 5^1\right)=180$

$ \begin{aligned} L=2^3 \cdot 3 \cdot 5 & =120, M=2 \cdot 3 \cdot 5^3=750 \\ \therefore \quad M-L & =2 \cdot 3 \cdot 5^3-2^3 \cdot 3 \cdot 5 \\ & =2 \cdot 3 \cdot 5(25-4) \\ & =2 \cdot 3 \cdot 5 \cdot 21=2 \cdot 3^2 \cdot 5 \cdot 7 \end{aligned} $

Since, $360=2^3 \cdot 3^2 \cdot 5^1$

The number of $(+\mathrm{ve})$ divisors which are multiple of 3

$ =(3+1)(2+1-1)(1+1)=4 \times 2 \times 2=16 $

The total possible arrangement of the word 'LINEAR' is $6!=720$.

Number of possible arrangement of the word 'LINEAR' in which $E$ and $A$ come together is $5!2!=120 \times 2=240$

Number of possible arrangement in the word 'LINEAR' in which $E$ and $A$ come together and $N$ and $R$ come together $=2!2!4!=2 \times 2 \times 24=96$

Number of possible arrangement in which $E$ and $A$ come together but $N$ and $R$ do not come together $=240-96=144$

Given, 15 lines are concurrent at a point $P$.

Now, if a line is not passing through a point $P$, then the triangle can be formed if those 15 times, concurrent at 2 points, so the required number of ways of forming a triangle $={ }^{15} C_2$

$ =\frac{15 \times 14}{1 \times 2}=15 \times 7=105 $

Case I When $x=1$, then $y+z=11$

There are 4 triplets in this case.

i.e. $(1,2,9),(1,3,8),(1,4,7),(1,5,6)$

Case II When $x=2$, then $y+z=10$

There are 2 triplets in this case i.e. $(2,3,7)$ and $(2,4,6)$.

Case III When $x=3$, then $y+z=9$

There is only one triplet in this case i.e. $(3,4,5)$.

∴ Total number of triplets $=4+2+1=7$

There are total 12 questions.

A candidate can choose 8 questions from 12 in the following pattern

$ \begin{array}{ccc} \hline \text { Ist } & \text { Ilnd } & \text { Illrd } \\ \hline 2 & 2 & 4 \\ \hline 2 & 3 & 3 \\ \hline 2 & 4 & 2 \\ \hline 3 & 2 & 3 \\ \hline 3 & 3 & 2 \\ \hline 4 & 2 & 2 \\ \hline \end{array} $

$ \begin{aligned} &\text { ∴ Total number of ways }\\ &\begin{aligned} = & \left({ }^4 C_2 \times{ }^4 C_2 \times{ }^4 C_4\right)+\left({ }^4 C_2 \times{ }^4 C_3 \times{ }^4 C_3\right) \\ & +\left({ }^4 C_2 \times{ }^4 C_4 \times{ }^4 C_2\right)+\left({ }^4 C_3 \times{ }^4 C_2 \times{ }^4 C_3\right) \\ & +\left({ }^4 C_3 \times{ }^4 C_3 \times{ }^4 C_2\right)+\left({ }^4 C_4 \times{ }^4 C_2 \times{ }^4 C_2\right) \\ = & 6 \times 6 \times 1+6 \times 4 \times 4+6 \times 1 \times 6+4 \times \\ & 6 \times 4+4 \times 4 \times 6+1 \times 6 \times 6 \\ = & 36+96+36+96+96+36=396 \end{aligned} \end{aligned} $

Total ways $=10$ !

Number of ways all three are together

$ =3!\times 8! $

∴ Required number of ways

$ =10!-3!\times 8!=84 \times 8! $

$6=2 \times 3$

Highest power of 2 in 72!

$ \begin{aligned} =\left[\frac{72}{2}\right]+\left[\frac{72}{2^2}\right]+\left[\frac{72}{2^3}\right]+\left[\frac{72}{2^4}\right] +\left[\frac{72}{2^5}\right]+\left[\frac{72}{2^6}\right] \end{aligned} $

$ \begin{aligned} &=36+18+9+4+2+1=70\\ &\text { Highest power of } 3 \text { in } 72 \text { ! }\\ &\begin{aligned} & =\left[\frac{72}{3}\right]+\left[\frac{72}{3^2}\right]+\left[\frac{72}{3^3}\right]=24+8+2=34 \\ & \therefore \text { Exponent of } 6 \text { in } 72!=34 \\ & \qquad\left(\because 2^{34} \times 3^{34}=6^{34}\right) \end{aligned} \end{aligned} $

$ \text { Taking } 1 \text { at one's place, } $

So, two numbers can be formed using the above combination.

So, two numbers can be formed using the above combination.

Total numbers that can be formed when 1 is at the unit's place

$ =2+2+2+2=8 $

Now, taking 3 at one's place,

Total numbers that can be formed when 3 is at the unit's place $=2+2+2+2=8$

Now, taking 5 at one's place,

Total numbers that can be formed when 5 is at the unit's place $=2+2+2+2=8$ ∴ Total number of 3 -digit odd numbers divisible by 3 that can be formed using the digits $1,2,3,4,5,6$ when repetition is not allowed $=8+8+8=24$

(A) The number of ways of not selecting $(n-r)$ things from $n$ different things $={ }^n C_{n-r}$

(B) $(n-r+1) \cdot{ }^n C_{r-1}$

$ \begin{aligned} & =(n-r+1) \times \frac{n!}{(r-1)!(n-r+1)!} \\ & =\frac{(n-r+1) \times n!}{(r-1)!(n-r+1)(n-r)!} \\ & =\frac{n!}{(r-1)!(n-r)!}=r \frac{n!}{r!(n-r)!}=r \cdot{ }^n C_r \end{aligned} $

(C) The number of ways of selecting atleast $(n-r)$ things from $n$ different things

$ \begin{aligned} & ={ }^n C_{n-r}+{ }^n C_{n-r+1}+{ }^n C_{n-r+2} +\ldots+{ }^n C_n \\ & ={ }^n C_n+{ }^n C_{n-1}+{ }^n C_{n-2}+\ldots \quad+{ }^n C_{n-r+1}+{ }^n C_{n-r} \end{aligned} $

$ \begin{aligned} & =1+{ }^n C_1+{ }^n C_2+\ldots+{ }^n C_r \\ & \text { (D) }(n-r)\left({ }^{n-1} C_{r-1}+{ }^{n-1} C_r\right) \\ & =(n-r)\left({ }^n C_r\right) \\ & {\left[\because{ }^n C_r+{ }^n C_{r+1}={ }^{n+1} C_{r+1}\right]} \\ & =(n-r) \times \frac{n!}{(n-r)!r!}=\frac{n!}{(n-r-1)!r!} \\ & =(r+1) \frac{n!}{(r+1)!(n-r-1)!}=(r+1){ }^n C_{r+1} \end{aligned} $

$10^n C_2=3^{n+1} C_3$

$\begin{aligned} & \frac{10 n!}{(n-2)!2!}=\frac{3(n+1)!}{(n+1-3)!\times 3!} \\ & \Rightarrow \quad \frac{10 n!}{2}=\frac{3(n+1) n!}{3 \times 2} \\ & \Rightarrow \quad 10=n+1 \\ & \Rightarrow \quad n=9 \\ \end{aligned}$

Here's how to solve this problem:

First, let's understand the concepts involved:

• Collinear points: Points that lie on the same straight line.
• Triangle: A polygon with three sides.

To form a triangle, we need to choose 3 points out of the 10. We can do this using combinations:

Total number of ways to choose 3 points out of 10 = $^{10}C_3 = \frac{10!}{3!7!} = 120$

However, this count includes triangles formed using the 6 collinear points. We need to subtract these invalid triangles.

Number of ways to choose 3 points out of 6 collinear points = $^6C_3 = \frac{6!}{3!3!} = 20$

Therefore, the total number of valid triangles (those not formed by the collinear points) is:

$N = 120 - 20 = 100$

So the answer is Option C: 100.

Total marks $=$ Marks for first 3 papers + Marks for fourth paper $=3 n+2 n=5 n$

Candidate needs to get $3 n$ marks.

Let $x_1, x_2, x_3, x_4$ be the marks of candidate in I, II, III and IV paper, respectively.

Then, $x_1+x_2+x_3+x_4 =3 n\left(0 \leq x_1, x_2, x_3, \leq n\right.$ and $\left.0 \leq x_4 \leq 2 n\right)$

Using multinomial theorem,

$\begin{aligned} & \begin{array}{l} \text { Number of ways }=\text { Coefficient of } x^{3 n} \text { in } \\ \underbrace{\left(1+x+\ldots+x^n\right)^3}_{\text {GP series }} \underbrace{\left(1+x+\ldots x^{2 n}\right)}_{\text {GP Series }} \end{array} \\ & \Rightarrow\left(\frac{1-x^{n+1}}{1-x}\right)^3\left(\frac{1-x^{2 n+1}}{1-x}\right) \\ & \Rightarrow 1-x^{3(n+1)}-3 x^{n+1}\left(1-x^{n+1}\right)\left(1-x^{2 n+1}\right)(1-x)^{-4} \\ & \Rightarrow\left(1-x^{3(n+1)}-3 x^{n+1}+3 x^{2 n+2}\right)\left(1-x^{2 n+1}\right)(1-x)^{-4} \\ & \Rightarrow\left(1-x^{3(n+1)}-3 x^{n+1}+3 x^{2 n+2}-x^{2 n+1}+x^{5 n+4}\right. \\ & \left.\quad+3 x^{3 n+2}-3 x^{4 n+3}(1-x)^{-4}\right) \end{aligned}$

Finding the coefficient of $x^{3 n}$ using binomial expansion, we get

$\begin{gathered} { }^{3 n+3} C_3-3^{2 n+2} C_3+3^{n+2} C_3-{ }^{n+3} C_3 \\ =\frac{1}{6}(n+1)\left(5 n^2+10 n+6\right) \end{gathered}$

If set $ A $ contains $ m $ elements and set $ B $ comprises $ n $ elements, the number of injections from $ A $ to $ B $ can be defined as follows:

$ \left\{ \begin{array}{cc} 0, & \text{if } n < m \\ {}^n P_m, & \text{if } n \geq m \end{array} \right. $

This means that if the number of elements in set $ B $ is less than the number in set $ A $ ($ n < m $), there are no injections possible. Conversely, if $ n \geq m $, the number of injections is given by the permutation notation $ {}^n P_m $.

MULTIPLE has 8 letters, with 5 consonant (L repeated twice) and 3 vowels.

The vowels in 2nd, 5th and 8th places, these can be arranged in 3 ! ways $=6$ ways

The remaining 5 consonants can be arranged in $\frac{5!}{2!}$ ways $=\frac{120}{2}=60$ ways

Thus, the total number of ways $=6 \times 60=360$

Number of zeroes at the end of 4000!

$\begin{aligned} & =\left[\frac{4000}{5}\right]+\left[\frac{4000}{5^2}\right]+\left[\frac{4000}{5^{-3}}\right]+\left[\frac{4000}{5^4}\right]+\left[\frac{4000}{5^5}\right] \\ & =800+160+32+6+1=999 \end{aligned}$

Now, we need one more pair of 5 and 2 so that the number ends in exactly 1000 zeroes.

$4005 \leq N<4009$

$\therefore$ From the option (c), $N=4009$

Total number of things $=n$

Repetition of things is allowed and atmost $r$ can be taken at a time.

The number of ways of taking 1 thing at a time $=n$

The number of ways of taking 2 things at a time $=n^2$

The number of ways of taking 3 things at a time $=n^3$ ..........

The number of ways of taking $r$ things at a time $=n^3$

Total number of permutations of $n$ different things taken not more than $r$ at a time i.e. number of permutations of taking $1,2, \ldots ., r$ at a time.

$\begin{aligned} & =n+n^2+n^3+\ldots .+n^{\prime} \\ & =\frac{n\left(n^2-1\right)}{n-1} \quad [\because ~\text{sum of GP}] \end{aligned}$

There are 21 points on the circle, denoted as $ n = 21 $. A chord is formed by connecting any two points on the circle.

Therefore, the number of possible chords is given by the combination formula $ { }^n C_2 $:

$ { }^{21} C_2 = \frac{21 \times 20}{2 \times 1} = 210 $

Hence, 210 chords can be drawn through the 21 points on the circle.

$\begin{aligned} & \text { Number of sides }=n \\ & \text { Number of diagonals }=560 \\ & \because \text { Number of diagonals }=\frac{n(n-3)}{2} \\ & \Rightarrow \quad 560=\frac{n(n-3)}{2} \\ & \Rightarrow \quad 1120=n^2-3 n \\ & \Rightarrow \quad n^2-3 n-1120=0 \\ & \Rightarrow \quad n^2-35 n+32 n-1120=0 \\ & \Rightarrow \quad n(n-35)+32(n-35)=0 \\ & \Rightarrow \quad(n-35)(n+32)=0 \\ & \Rightarrow \quad n=35,-32 \\ & \qquad n \neq-32 \\ & \therefore \quad n=35. \end{aligned}$

If the total number of letters is $n$, then the number of ways in which $r$ letters goes into wrong envelopes $={ }^n C_{n-r} D_r$

The number of ways in which atleast two of the letters are in wrong envelopes

$\begin{aligned} & ={ }^6 C_{6-2} D_2+{ }^6 C_{6-3} D_3+{ }^6 C_{6-4} D_4 \\ & +{ }^6 C_{6-5} D_5+{ }^6 C_{6-6} D_6 \quad[\because r=2,3,4,5,6] \\ & =\sum_{r=2}^6{ }^6 C_{6-}, D_r \quad\left[\text { where, } D^r=r!\left(\sum_{i=0}^r \frac{(-1)^r}{i!}\right)\right] \\ \end{aligned}$

Other than 2, remaining five places can be filled by 1 and 3 for each place. The number of ways for five places is 2 $\times$ 2 $\times$ 2 $\times$ 2 $\times$ 2 = 2⁵. For 2, selecting 2 places out of 7 is ⁷C₂. Hence, the required number of ways is ⁷C₂ $\times$ 2⁵.

The word HAVANA has 6 letters. However, the letters 'A' appear 3 times. We'll first consider all permutations, and then subtract the ones where 'V' and 'N' appear together.

1. Total permutations :

Since we have 6 total letters and 3 of them are the same (A), we use the formula for permutations of multiset to get the total number of arrangements, which is 6! / 3! = 720 / 6 = 120.

2. Permutations with 'V' and 'N' together :

Let's consider 'V' and 'N' as a single entity. So, we have 5 entities - {V,N}, 'A', 'A', 'A', 'H'.

Since 'A' appears 3 times, the total permutations would be 5! / 3! = 120 / 6 = 20.

However, 'V' and 'N' can be arranged among themselves in 2! = 2 ways.

So, total arrangements where 'V' and 'N' are together is 20 $ \times $ 2 = 40.

The total permutations where 'V' and 'N' do not appear together is the total permutations minus the ones where they are together: 120 - 40 = 80.

Total number of triangles = ${}^{15}{C_3}$

i + j + k = 15 (Given)


Number of possible triangles using the vertices P_i, P_j, P_k such that i + j + k $\ne$ 15 is equal to ${}^{15}{C_3}$ $-$ 12 = 443

Option (c)

${}^n{P_r} = {}^n{P_{r + 1}} \Rightarrow {{n!} \over {(n - r)!}} = {{n!} \over {(n - r - 1)!}}$

$ \Rightarrow (n - r) = 1$ .....(1)

${}^n{C_r} = {}^n{C_{r - 1}}$

$ \Rightarrow {{n!} \over {r!(n - r)!}} = {{n!} \over {(r - 1)!(n - r + 1)!}}$

$ \Rightarrow {1 \over {r(n - r)!}} = {1 \over {(n - r + 1)(n - r)!}}$

$ \Rightarrow n - r + 1 = r$

$ \Rightarrow n + 1 = 2r$ ..... (2)

From (1) and (2), $ 2r - 1 - r = 1 \Rightarrow r = 2$

Total possible numbers using 1, 2, 2 and 3 is

= ${{4!} \over {2!}}$ = 12

When unit place is 1, the total possible numbers using remaining 2, 2 and 3 are

= ${{3!} \over {2!}}$ = 3

When unit place is 2, the total possible numbers using remaining 1, 2 and 3 are

= 3! = 6

When unit place is 3, the total possible numbers using remaining 1, 2 and 2 are

= ${{3!} \over {2!}}$ = 3

$ \therefore $ Sum of unit places of all (3 + 6 + 3) 12 numbers is

= ( 1$ \times $3 + 2$ \times $6 + 3$ \times $3)

Similarly,

When 10^th place is 1, the total possible numbers using remaining 2, 2 and 3 are

= ${{3!} \over {2!}}$ = 3

When 10^th place is 2, the total possible numbers using remaining 1, 2 and 3 are

= 3! = 6

When 10^th place is 3, the total possible numbers using remaining 1, 2 and 2 are

= ${{3!} \over {2!}}$ = 3

$ \therefore $ Sum of 10^th places of all (3 + 6 + 3) 12 numbers is

= ( 1$ \times $3 + 2$ \times $6 + 3$ \times $3) $ \times $ 10

Similarly,

Sum of 100^th places of all (3 + 6 + 3) 12 numbers is

= ( 1$ \times $3 + 2$ \times $6 + 3$ \times $3) $ \times $ 100

and Sum of 1000^th places of all (3 + 6 + 3) 12 numbers is

= ( 1$ \times $3 + 2$ \times $6 + 3$ \times $3) $ \times $ 1000

$ \therefore $ Total sum = ( 1$ \times $3 + 2$ \times $6 + 3$ \times $3) + ( 1$ \times $3 + 2$ \times $6 + 3$ \times $3) $ \times $ 10

+ ( 1$ \times $3 + 2$ \times $6 + 3$ \times $3) $ \times $ 100 + ( 1$ \times $3 + 2$ \times $6 + 3$ \times $3) $ \times $ 1000

= (3 + 12 + 9) (1 + 10 + 100 + 1000) = 1111 $ \times $ 24 = 26664

Total number of triangles

= ${}^{14}{C_3} - {}^3{C_3} - {}^5{C_3} - {}^6{C_3}$

= 364 – 31 = 333

Total matches between boys of both team = ${}^7{C_1} \times {}^4{C_1} = 28$

Total matches between girls of both team = ${}^n{C_1}\,{}^6{C_1} = 6n$

Now, 28 + 6n = 52

$ \Rightarrow $ n = 4

$\alpha = {}^6{C_1}{}^7{C_1}{}^9{C_1} + {}^5{C_1}{}^7{C_1}{}^9{C_1} + {}^5{C_1}{}^6{C_1}{}^9{C_1} + {}^5{C_1}{}^6{C_1}{}^7{C_1} $

$= 378 + 315 + 270 + 210 = 1173$

$\beta = {}^5{C_1}{}^6{C_1}{}^7{C_1}{}^9{C_1} = 1890$

$ \therefore $ $ \beta - \alpha = 1890 - 1173 = 717$

y + z = 5 ....... (1)

${1 \over y} + {1 \over z} = {5 \over 6}$

$ \Rightarrow {{y + z} \over {yz}} = {5 \over 6}$

$ \Rightarrow {5 \over {yz}} = {5 \over 6}$

$ \Rightarrow $ yz = 6

Also, (y $-$ z)² = (y + z)² $-$ 4yz

$ \Rightarrow $ (y $-$ z)² = (y + z)² $-$ 4yz

$ \Rightarrow $ (y $-$ z)² = 25 $-$ 4(6) = 1

$ \Rightarrow $ y $-$ z = 1 ..... (2)

from (1) and (2), y = 3 and z = 2

for calculating odd divisor of p = 2^x . 3^y . 5^z

x must be zero

P = 2⁰ . 3³ . 5²

$ \Rightarrow $ Total possible cases = (3⁰5⁰ + 3¹5⁰ + 3²5⁰ + 3³5⁰ + .... + 3³5²)

$ \therefore $ Total odd divisors must be (3 + 1) ( 2 + 1) = 12

(I) First possibility is 1, 1, 1, 1, 1, 2, 3

required number = ${{7!} \over {5!}}$ = 7 $\times$ 6 = 42

(II) Second possibility is 1, 1, 1, 1, 2, 2, 2

required number = ${{7!} \over {4!3!}} = {{7 \times 6 \times 5} \over 6} = 35$

Total = 42 + 35 = 77

$x.y.z = 24$

$x.y.z = {2^3}.\,{3^1}$

Three 2 has to be distributed among x, y and z

Each may receive none, one or two

$\therefore$ Number of ways = ${}^{3 + 3 - 1}{C_{3 - 1}}$ = $^5{C_2}$ ways

Similarly one 3 has to be distributed among x, y and z

$ \therefore $ Number of ways = ${}^{1 + 3 - 1}{C_{3 - 1}}$ = $^3{C_2}$ ways

Total ways = $^5{C_2}\,.{\,^3}{C_2}$ = 30