2 different vowels and 2 different consonants are to be selected from the English alphabet.
Since there are 5 vowels in the English alphabet, number of ways of selecting 2 different vowels from the alphabet \(={ }^{5} \mathrm{C}_{2}=\frac{5 !}{2 ! 3 !}=10\)
Since there are 21 consonants in the English alphabet, number of ways of selecting 2 different consonants from the alphabet 
\(={ }^{21} \mathrm{C}_{2}=\frac{21 !}{2 ! 19 !}=210\)
=\(\frac { 5\times 4 }{ 2 } \frac { 21\times 20 }{ 2 } \times 24\)
Therefore, number of combinations of 2 different vowels and 2 different consonants = 10 × 210 = 2100
Each of these 2100 combinations has 4 letters, which can be arranged among themselves in 4! ways.
Therefore, required number of words = 2100 × 4! = 50400

Let us first seat the 5 girls. This can be done in 5! ways. For each such arrangement, the three boys can be seated only at the cross marked places.
× G × G × G × G × G ×.
There are 6 cross marked places and the three boys can be seated in ⁶P₃ ways.
Hence, by multiplication principle, the total number of ways 
\(=5 ! \times{ }^{6} \mathrm{P}_{3}=5 ! \times \frac{6 !}{3 !}\)
= 4 × 5 × 2 × 3 × 4 × 5 × 6 = 14400.

5 men and 4 women are to be seated in a row such that the women occupy the even places.
The 5 men can be seated in 5! ways. For each arrangement, the 4 women can be seated only at the cross marked places (so that women occupy the even places).
M x M x M x M x M
Therefore, the women can be seated in 4! ways.
Thus, possible number of arrangements = 4! × 5! = 24 × 120 = 2880

Here we have to choose 10 student out of a class 25 , such that the given condition satisfied
let us make the following cases anf find the number of paossible choices in each case
Case I when 3 particular sudents join the party
In this case, we have to choose 7 student out of 22 students. This can be done in \(^{ 27 }{ C }{ _{ 7 } }\)ways
Case II when 3 particular sudents join the party
In this case, we have to choose 10 student out of 22 students. This can be done in \(^{ 22 }{ C }{ _{ 10 } }\) ways
Hence, total number of ways in which excurson party can be chosen =\( ^{ 22 }{ C }{ _{ 7 } } + ^{ 22 }{ C }{ _{ 10 } }\)

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.

Since, 1000000 is a 7-digit number and the number of digits to be used is also 7. Therefore, the numbers to be counted will be 7-digit only. Also, the numbers have to be greater than 1000000, so they can begin either with 1, 2 or 4.
The number of numbers beginning with 1 \(=\frac{6 !}{3 ! 2 !}=\frac{4 \times 5 \times 6}{2}=60\) as when 1 is fixed at the extreme left position, the remaining digits to be rearranged will be 0, 2, 2, 2, 4, 4, in which there are 3, 2s and 2, 4s.
Total numbers begining with 2 \(=\frac{6 !}{2 ! 2 !}=\frac{3 \times 4 \times 5 \times 6}{2}=180\) and total numbers begining with 4 \(=\frac{6 !}{3 !}=4 \times 5 \times 6=120\) Therefore, the required number of numbers = 60 + 180 + 120 = 360.
Alternative Method :
The number of 7-digit arrangements, clearly \(\frac{7 !}{3 ! 2 !}=420\) But, this will include those numbers also, which have 0 at the extreme left position. The number of such arrangements \(\frac{6 !}{3 ! 2 !}\) (by fixing 0 at the extreme left position) = 60.
Therefore, the required number of numbers = 420 – 60 = 360.

Out of 17 players, 5 players are bowlers.
A cricket team of 11 players is to be selected in such a way that there are exactly 4 bowlers.
4 bowlers can be selected in  ⁵C₄ ways and the remaining 7 players can be selected out of the 12 players in ¹²C₇ ways.
Thus, by multiplication principle, required number of ways of selecting cricket team
= ⁵C₄ x ¹²C₇
\(={5!\over 4!1!}\times{12!\over 7!5!}\)
\(={5\times4!\over 4!}\times{12\times11\times10\times9\times8\times7!\over 7!\times5\times4\times3\times2\times1}\)
= 5 x 792 = 3960

When a coin is tossed once, the number of outcomes is 2 (Head and tail) i.e., in each throw, the number of ways of showing a different face is 2.
Thus, by multiplication principle, the required number of possible outcomes is 2 × 2 × 2 = 8

(i) There will be as many ways as there are ways of filling 3 vacant places  in succession by the given five digits. In this case, repetition of digits is allowed. Therefore, the units place can be filled in by any of the given five digits. Similarly, tens and hundreds digits can be filled in by any of the given five digits.Thus, by the multiplication principle, the number of ways in which three-digit numbers can be formed from the given digits is 5 × 5 × 5 = 125
(ii) In this case, repetition of digits is not allowed. Here, if units place is filled in first, then it can be filled by any of the given five digits. Therefore, the number of ways of filling the units place of the three-digit number is 5. Then, the tens place can be filled with any of the remaining four digits and the hundreds place can be filled with any of the remaining three digits. Thus, by the multiplication principle, the number of ways in which three-digit numbers can be formed without repeating the given digits is 5 × 4 × 3 = 60

(i) Here ²ⁿC₃ : ⁿC₂ = 12 : 1
\(⇒\ {(2n)!\over 3!(3n-3)}\times{2(n-2)!\over n !}={12\over 1}\)
\(⇒\ {(2n)(2n-1)(2n-)2n-22)(2n-3)!\over 3\times2!(2n-3)!}\times{2!(n-2)!\over n(-1)(n-2)!}={12\over 1}\)
\(⇒\ {(2n)(2n-1)(2n-2)\over 3}\times{1\over n(n-1)}={12\over 1}\)
\(⇒\ {4(2n-1)\over 3}={12\over1}\)
⇒ 8n - 4 = 36 ⇒ n = 5
(ii) Here ²ⁿc₃: ⁿC₃ = 11 : 1
\(⇒\ {(2n)!\over 3!(2n-3!)}\times{2!(n-3)!\over n!}={11\over 1}\)
\(⇒\ {(2n)(2n-1)(2n-2)(2n-3)!\over 3!(2n-3)!}\times{3!(n-3)!\over n(n-1)(n-2)(n-3)!}={11\over 1}\)
\(⇒ {(2n)(2n-1)(2n-2)\over n(n-1)(n-2)}={11\over 1}\)
\(⇒\ {2(2n-1)\over n-2}={11\over 1}\)
⇒ 8n - 4 = 11n - 22
⇒ 3n = 18 ⇒ n = 6.

Proof is very similar to that of Theorem 1 and is left for the reader to arrive at. Here, we are solving some of the problems of the pervious Section using the formula for nPr to illustrate its usefulness. In Example 1, the required number of words = ⁴P₄ = 4! = 24. Here repetition is not allowed. If repetition is allowed, the required number of words would be 4⁴ = 256. The number of 3-letter words which can be formed by the letters of the word NUMBER \(={ }^6 \mathrm{P}_3=\frac{6 !}{3 !}=4 \times 5 \times 6=120\). Here, in this case also, the repetition is not allowed. If the repetition is allowed, the required number of words would be \(6^3=216\). The number of ways in which a Chairman and a Vice-Chairman can be chosen from amongst a group of 12 persons assuming that one person can not hold more than one position, clearly \({ }^{12} \mathrm{P}_2=\frac{12 !}{10 !}=11 \times 12=132\)

There are 12 letters, of which N appears 3 times, E appears 4 times and D appears 2 times and the rest are all different. Therefore
The required number of arrangements \(=\frac{12 !}{3 ! 4 ! 2 !}=1663200\)
(i) Let us fix P at the extreme left position, we, then, count the arrangements of the remaining 11 letters. Therefore, the required number of words starting with P
\(=\frac{11 !}{3 ! 2 ! 4 !}=138600\)
(ii) There are 5 vowels in the given word, which are 4 Es and 1 I. Since, they have to always occur together, we treat them as a single object EEEEI for the time being. This single object together with 7 remaining objects will account for 8 objects. These 8 objects, in which there are 3Ns and 2 Ds, can be rearranged in \(\frac{8 !}{3 ! 2 !}\) ways. Corresponding to each of these arrangements, the 5 vowels E, E, E, E and I can be rearranged in \(\frac{5 !}{4 !}\) ways. Therefore, by multiplication principle, the required number of arrangements
\(=\frac{8 !}{3 ! 2 !} \times \frac{5 !}{4 !}=16800\)
(iii) The required number of arrangements = the total number of arrangements (without any restriction) – the number of arrangements where all the vowels occur together.
= 1663200 – 16800 = 1646400
(iv) Let us fix I and P at the extreme ends (I at the left end and P at the right end). We are left with 10 letters. Hence, the required number of arrangements
\(=\frac{10 !}{3 ! 2 ! 4 !}=12600\)

Total number of discs are 4 + 3 + 2 = 9. Out of 9 discs, 4 are of the first kind (red), 3 are of the second kind (yellow) and 2 are of the third kind (green).
Therefore, the number of arrangements \(\frac{9 !}{4 ! 3 ! 2 !}=1260\).

(i) There are 8 different letters in the word DAUGHTER, in which there are 3 vowels, namely, A, U and E. Since the vowels have to occur together, we can for the time being, assume them as a single object (AUE). This single object together with 5 remaining letters (objects) will be counted as 6 objects. Then we count permutations of these 6 objects taken all at a time. This number would be 6P6 = 6!. Corresponding to each of these permutations, we shall have 3! permutations of the three vowels A, U, E taken all at a time . Hence, by the multiplication principle the required number of permutations = 6 ! × 3 ! = 4320.
(ii) If we have to count those permutations in which all vowels are never together, we first have to find all possible arrangments of 8 letters taken all at a time, which can be done in 8! ways. Then, we have to subtract from this number, the number of permutations in which the vowels are always together.
Therefore, the required number 8 ! – 6 ! × 3 ! = 6 ! (7×8 – 6)
= 2 × 6 ! (28 – 3)
= 50 × 6 ! = 50 × 720 = 36000