When atleast 3 girls are included, then
Number of ways = ⁴C₃ \(\times\)⁷C₂ + ⁴C₄ \(\times\)⁷C_I
\(=4\times {7\times6\over2\times1}+1\times7=84+7=91ways\)
Hence the required number of ways are 91 ways

When at least one by and one girl are to be selected, then
Number of ways=⁴C₁ \(​​\times\) ⁷C₄ +⁴C₇\(​​\times\) ⁷C₃ +⁴C₃\(​​\times\)⁷C₂ +⁴C₄+ ⁷C₁
\(=4\times{7\times6\times5\times4\over4\times3\times2\times1}+{4\times3\over2\times1}\times{7\times6\times5\over3\times2\times1}+4\times{7\times6\over2\times1}+1\times7\)
= (4 \(​​\times\) 35) + (6 \(​​\times\) 35) + (4 \(​​\times\)21) + 7
= 140 + 120 + 84 + 7 = 441 ways
Hence the required number of ways are 441 ways​​​​​​​

We have 4 girls and 7 boys and a team of 5 members is to be selected.
If no girl in selected, then all the 5 members are to be selected out of 7 boys
(i.e) ⁷C₅ = \({7!\over 5!2!}={7\times 6.5!\over5!\times2}=21ways\) 
Hence the required number of ways are 21 ways

3 , 2

4

We have nCx = ⁿCy ⇒ x = y or x + y = n
⇒ 12 + 9 = n
⇒ n = 21
⇒ 21C_n = ²¹C₂₁ 
= 1     [∵ ⁿC_n= 1]

Given \(\frac { 2n{ C }_{ 3 } }{ n{ C }_{ 3 } } \) = \(\frac { 11 }{ 1 } \)
\(\Rightarrow\)  2nC₃ = 11.nC₃

\(\Rightarrow\) 8n - 4 = 11n - 22 \(\Rightarrow\) 4n
\(\Rightarrow\) 22-4 = 3n
\(\Rightarrow\) 3n = 18 \(\Rightarrow\) n = 6.

There are 6 periods in each working day of a school.
Since each subject is allowed at least one period, we first select one subject for the left out period. This can be done in 5C₁ ways.
Now, six subject can be arranged in \(=\frac { 6! }{ 2! } \) ways
Hence, total number of permutations = 5C₁ \(\times\)\(\frac { 6! }{ 2! } \)ways

= 1800.

4 coats can be given to 3 men in 4 p₃ ways.5 waist coats can be given to 3 men in 5P₃ ways 6 caps can be given to 3 men in 6 P₃ ways.
∴ Total number of ways of wearing them 
= 4P3 \(\times\) 5P₃ \(\times\) 6 P₃ 
= \(\frac { 4! }{ 1! } \times \frac { 5! }{ 2! } \times \frac { 6! }{ 3! } \)
= \(4\times 3\times 2\times \frac { 5\times 4\times 3\times 2! }{ 2! } X\frac { 6\times 5\times 4\times 3! }{ 3! } \)
= 24 \(\times\) 60 \(\times\) 120 
= 1,72,800

Gold medal can be awarded to anyone of the 8 candidates in 8 ways.
Silver medal can be awarded to anyone of the remaining 7 candidates in 7 ways.
Bronze medal can be awarded to anyone of the remaining 6 candidates in 6 ways.
∴ Total numbers of ways of awarding the prize
= 8 \(\times\) 7 \(\times\) 6 = 336

= \(\frac { 12\times 11\times 10\times 9! }{ 9!\quad 3\times 2 } =\frac { 12\times 11\times 10 }{ 6 } \)
= 2 \(\times\) 11 \(\times\) 10 = 220

Since each question can be answered in 4 ways, the total number of ways of answering 6 questions is
 \(4 \times 4 \times 4 \times 4 \times 4 \times 4=4^{6}\)

LHS =¹⁵C₃ + 2 \(\times\) ¹⁵C₄ +¹⁵C₄ +¹⁵C₅
= (¹⁵C₃ +¹⁵C₄) + (¹⁵C₄ + ¹⁵C₅)
=(¹⁵C₃ +¹⁵C₄) + (¹⁵C₄ + ¹⁵C₅)       [∴ ⁿC_r-1 + ⁿC_r = n +¹C_r]
= ¹⁶C₄ + (¹⁵C₄ +¹⁵C₅)
=¹⁶C₄ + ¹⁶C₅
= ¹⁷C₅ = RHS.

To form a triangle we need minimum 3 non-collinear points.
points can be selected from 15 non-collinear points in 15C₃ ways.
= \(\frac { 15\times 14\times 13 }{ 3\times 2\times 1 } \)
= 455

Examination has 11 letters, and in which 'A' 'I' and 'N', all occur twice 11C4' would have been fine if all letters were distinct. So we have E, X, M, T, 0, (AA), (II), (NN). 8 distinct letters.
\( { }^{8} \mathrm{C}_{4} \times 4 ! =\frac{8 \times 7 \times 6 \times 5}{1 \times 2 \times 3 \times 4} \times 4 \times 3 \times 2 \times 1 \)
\( =1680 \)
\({ }^{3} C_{1} \times{ }^{7} C_{2} \times \frac{4 !}{2 !} =3 \times \frac{7 \times 6}{1 \times 2} \times \frac{4 \times 3 \times 2 \times 1}{2 \times 1}=756 \)
\({ }^{3} C_{2} \times \frac{4 !}{2 ! 2 !} =\frac{3 \times 2}{1 \times 2} \times \frac{4 \times 3 \times 2 \times 1}{2 \times 2}=18 \)
Total no. of ways = 2454