When a coin is tossed, the outcomes are in two ways which are {Head, Tail}.
By the rule of product rule, the number of outcomes when 5 coins are tossed is \(2 \times 2 \times 2 \times 2 \times 2=2^{5}=32\)

\(5 !=5 \times 4 \times 3 \times 2 \times 1=120\)

\(\frac { 8! }{ 5!\times 2! } =\frac { 8\times 7\times 6\times 5! }{ 5!\times 2! } =\frac { 8\times 7\times 6 }{ 2 } \) = 168

When n = 7, r = 5
\(\frac { n! }{ r!(n-r)! } =\frac { 7! }{ 5!(7-5)! } =\frac { 7\times 6\times 5! }{ 5!\times 2! } =\frac { 7\times 6 }{ 1\times 2 } \) = 21

\(\frac { 6! }{ n! } =\frac { 1.2.3.4.5.6 }{ 1.2.3...n } \) = 6. As n < 6 we get, n = 5

¹⁰C₃ = \(\frac { 10! }{ 7!\times 3! } =\frac { 10\times 9\times 8\times 7! }{ 7!\times 3\times 2\times 1 } \)
= \(\frac { 10\times 9\times 8 }{ 3\times 2\times 1 } \) = 120

¹⁰⁰C₉₉ = \(\frac { 100\times 99! }{ 99! } \) =100

There are 8 girls and 7 boys in the mathematics club.
The number of ways of selecting 6 members in that half of them girls (3 girls and 3 others) is ⁸C₃ \(\times\) ⁷C₃ = 56 \(\times\) 35 = 1960.

The first task of selecting a girl can be done in 29 ways. The second task of selecting a boy can be done in 17 ways. It follows from the sum rule, that there are 17 + 29 = 46 ways of making this selection.

There are 2 roads connecting Chennai to Trichy. Suppose these are R₁ and R₂. Further, there are 3 roads connecting Trichy to Tirunelveli. Let us name them as S₁, S₂ and S₃. Suppose a person chooses R₁ to travel from Chennai to Trichy and may further choose any of the 3 roads S₁, S₂ or S₃ to travel from Trichy to Tirunelveli. Thus the possible road choices are (R₁, S₁), (R₁, S₂), (R₁, S₃). Similarly, if the person chooses R₂ to travel from Chennai to Trichy, the choices would be (R₂, S₁), (R₂, S₂), (R₂, S₃).

Thus there are 2 \(\times\) 3 = 6 ways of travelling from Chennai to Tirunelveli.