n=6

Ans. 4320 

Number of ways of awarding scholarship for Classical language = 3
Number of ways of awarding scholarship for Mathematics = 5
Number of ways of awarding scholarship for Natural Science = 4.
Ans. (i) 60
(ii) 12

Here ⁿP_r  = ⁿP_r+1
\(\Rightarrow \frac { n! }{ (n-r)! } =\frac { n! }{ (n-r-1)! } \)
\(\Rightarrow \frac { 1 }{ (n-r)(n-r-1)! } =\frac { 1 }{ (n-r-1)! } \)
\(\Rightarrow \frac { 1 }{ n-r } =1 \Rightarrow n-r=1\) ....(i)
Also ⁿC_r = ⁿC_r-1
\(\Rightarrow \frac { n! }{ (n-r)!r! } =\frac { n! }{ (n-r+1)!(r-1)! } \)
\(\Rightarrow \frac { 1 }{ (n-r)!(r-1)! } =\frac { 1 }{ (n-1+1)(n-r)!(r-1)! } \)
\(\Rightarrow \frac { 1 }{ r } =\frac { 1 }{ n-r+1 } \)
\(\Rightarrow \)  n-r+1=r \(\Rightarrow \) n - 2r = -1  ....(ii)
From (i) and (ii), we have
n = 3 and r = 2.

A number is divisible by 10 if its units digits is 0.
Therefore, 0 is fixed at the units place.
Therefore, there will be as many ways as there are ways of filling 5 vacant places  
in succession by the remaining 5 digits (i.e., 1, 3, 5, 7 and 9).
The 5 vacant places can be filled in 5! ways.
Hence, required number of 6-digit numbers = 5! = 120

There are three servants to carry the cards. So the number of ways of sending the invitation card to the friends are 3.
So there are 3 ways of sending the invitation card to each of the five friends.
Thus required number of ways
= 3 \(\times\) 3 \(\times\)3 \(\times\)3 \(\times\) 3
= 3⁵ = 243

In the given word ASSASSINATION, the letter A appears 3 times, S appears 4 times, I appears 2 times, N appears 2 times, and all the other letters appear only once.
Since all the words have to be arranged in such a way that all the Ss are together, SSSS is treated as a single object for the time being. This single object together with the remaining 9 objects will account for 10 objects.
These 10 objects in which there are 3 As, 2 Is, and 2 Ns can be arranged in \(=\frac { 10! }{ 3!2!2! }\) ways
Thus, required number of ways of arranging the letters of the given word
\(=\frac { 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3! }{ 3!2\times 1\times 2\times 1 } \)
= 10 \(\times\) 9 \(\times\) 8 \(\times\)7 \(\times\) 6\(\times\) 5 = 151200.

\(\frac { 1 }{ 6! } +\frac { 1 }{ 7! } =\frac { x }{ 8! } \)
\(\Rightarrow \ \frac { 1 }{ 6! } +\frac { 1 }{ 7\times 6! } =\frac { x }{ 8\times 7\times 6! } \)
\(\Rightarrow \ \frac { 1 }{ 6! } [1+\frac { 1 }{ 7 } ] \ =\frac { 1 }{ 6! } [\frac { x }{ 8\times 7 } ]\)
\(\Rightarrow \ \frac { 8 }{ 7 } =\frac { x }{ 8\times 7 } \ \Rightarrow \ x=64\)

\(\frac { 8! }{ 6!\times 2! } =\frac { 8\times 7\times 6! }{ 6!\times 2! } =\frac { 8\times 7 }{ 2 } =28\)

Since each letter can be posted in any one of  4 letter boxes, so one letter can be posted in 4 ways. There are total 5 letters. So total number of ways =4x4x4x4x4
Answer:1024

(i) 60 (ii) 108

A team of 3 boys and 3 girls is to be selected from 5 boys and 4 girls.
3 boys can be selected from 5 boys in ⁵C₃ ways.
3 girls can be selected from 4 girls in ⁴C₃ ways.
Therefore, by multiplication principle, number of ways in which a team of 3 boys  and 3 girls can be selected
= ⁵C₃ x ⁴C₃
\(={ }^{5} \mathrm{C}_{3} \times{ }^{4} \mathrm{C}_{3}=\frac{5 !}{3 ! 2 !} \times \frac{4 !}{3 ! 1 !}\)
\(={5!\over 3!2!}\times{4!\over 3!1!}=10\times4=40\)

Here total number of digits= 9
Number of digits used (no digit is repeated)= 3
∴ Number of permutationa= ⁹P₃
\(={9!\over 6!}={9\times8\times7\times6!\over 6!}=504\)

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