\({1\over 7!}+{1\over 8!}={x\over 9!}\)
⇒ \({1\over 7!}+{1\over 8\times7!}={x\over 9\times8\times7!}\)
⇒ \({1\over 7!}\left[1+{1\over8}\right]={1\over 7!}\left[x\over 9\times8\right]\)
⇒ \({9\over 8}={x\over 9\times8}\)
⇒ x=81

Required number of signals is same as number of permutatons of 8 different things taken 5 at a time.
Ans. 6720

100

We have ,  ⁸C_r   - ⁷C₃ = ⁷C₂ 
\(\Longrightarrow \) ⁸C_r   =  ⁷C₃ + ⁷C₂ 
\(\Longrightarrow \) ⁸C_r   =  ⁸C₃      [\(\because\)  ⁿC_r+ ⁿC_r-1 = ⁿ⁺¹C_r]
\(\Longrightarrow \)r = 3 or r + 3 = 8 [ \(\because\) ⁿC_x = ⁿC_y  \(\Longrightarrow\) x = y or x + y = n ]
\(\Longrightarrow \)r =3 or r = 5

A man many invite one of them , two of them , three of them .... or all of them and this can be done in \(^{7}{C}{_1}, ^{7}{C}{_2}, ^{7}{C}{_3},...., ^{7}{C}{_7}\)
\(\because \) total number of ways
=\(^{7}{C}{_1}+ ^{7}{C}{_2}+ ^{7}{C}{_3}+^{7}{C}{_4}+^{7}{C}{_5}+^{7}{C}{_6}+^{7}{C}{_7}\)
=7+21-35-+35+21+7+1=127

We have, \(7!=7\times 6!\) and \(8!=8\times 7\times 6!\)
\(\therefore \) LCM of \(6!,7!\) and \(8!\) \(=LCM(6!,7\times 6!,8\times 7\times 6!)\)
\(=6!\times 7\times 8=8\times 7\times 6!=8!\)

Let total number of perons be n
Since, total number of hand-shakes=66
\(\therefore \)  \( ^{ n }{ C }{ _{ 2 } }=\)66  \(\Rightarrow\) \(\frac { n(n-1) }{ 2 } =66\)
\(\Rightarrow\) \( { n }^{ 2 }-n-132=0\)  \(\Rightarrow\)  (n-12)(n+11)=0
\(\therefore \) n=12 [ \(\because\)  cannot be negative ]

 \(^{ 14 }{ C }{ _{ 3 } }= \frac { 14! }{ 3!(14-3)! } \left[ \because \quad ^{ n }{ C }{ _{ r } }= \frac { n! }{ r!(n-r)! } \right] \)
 \(= \frac { 14! }{ 3!(14-3)! } = \frac { 14\times 13\times 12\times 11!\quad }{ (3\times 2\times 1)\times 11! } \)
 \(= \frac { 14! }{ 3!(14-3)! } = \frac { 14\times 13\times 12\times 11!\quad }{ \quad (3\times 2\times 1)\times 11! }\)
\(= 14 \times 13 \times 2\)       
=  364

0 cannot be placed at thousand's place. So, thousand's place can be filled in 9 ways. Since repitition of digits is allowed, therefore each of the remaining 3 places can be filled in 10 ways.
Ans. 9000

There will be as many ways as there are ways of filling 2 vacant places  in succession by the five given digits. Here, in this case, we start filling in unit’s place, because the options for this place are 2 and 4 only and this can be done in 2 ways; following which the ten’s place can be filled by any of the 5 digits in 5 different ways as the digits can be repeated. Therefore, by the multiplication principle, the required number of two digits even numbers is 2 × 5, i.e., 10.

Required number of ways = ³⁵P₂

Required number of ways = ⁷C₃ x ⁵C₂

7200

Ans. 840; 20 

Required numbers will be of 1-digit, 2-digit or 3-digit.
Ans. 155

Total letters in the word PERMUTATIONS = 12.
Here T = 2.
(i) Now first letter is P and last letter is S, which are fixed.
So the remaining 10 letters are to be arranged between P and S
∴ Number of permutations = \(10!\over 2!\)
\(={10\times9\times8\times7\times6\times5\times4\times3\times2!\over 2!}\)
=1814400
(ii) There are 5 vowels in the word PERMUTATIONS. All vowels can be put together.
∴ Number of permutations of all vowels together = ⁵p₅
\(={5!\over 0!}=5 \times 4 \times 3 \times 2 \times 1 = 120\)
Now consider the 5vowels together as one letter. Sothe number of letters in the word when all vowels are together = 8.
∴ Number of permutations  \(={8!\over 2!}\)
\(={8\times7\times6\times5\times4\times3\times2!\over 2!}\)
=20160
Hence the total number of permutations
= 120 x 20160 = 2419200
(iii) Here P and S are on 1st and 6th places
P and S are on 2nd and 7th places
P and S are on 3rd and 8th places P and S are on 4th and 9th places
P and S are on 5th and 10th places P and S are on 6th and 11th places
P and S are on 7th and 12th places
Now we see that P and S can be put in 7 ways and also P and S can interchange their positions.
∴ Number of permutations = 2 x 7 = 14
Now the remaining 10 places can be filled with remaining 10 letters.
∴ Number of permutations =  \(10!\over 2!\)
\(={10\times9\times8\times7\times6\times5\times4\times3\times2!\over 2!}\)
=1814400
Thus total number of permutations = 14 x 1814400= 25401600