8 litres are drawn from a cask full of wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of water is 16 : 65 . How much wine did the cask hold originally? a.18 litres b.24 litres c.32 litres d.42 litres

Answer:

c. opposite sides are parallel.

Step-by-step explanation:

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Answer:

c) opposite sides are parallel

Answer:

10 days

Step-by-step explanation:

because A does 5 days more fast work than b

Answer:

A simple event is one that can only happen in one way - in other words, it has a single outcome. If we consider our previous example of tossing a coin: we get one outcome

Answer:

Simple events are the events where one experiment happens at a time and it will be having a single outcome. The probability of simple events is denoted by P(E) where E is the event. The probability will lie between 0 and 1. For example, tossing of coin is a simple event.

Step-by-step explanation:

We know, from definition of binomial expansion,

[tex]\sf \: {(1 + x)}^{n} = C_0 + C_1x + C_2 {x}^{2} + ... + C_{n - 1} {x}^{n - 1} + C_n {x}^{n} \\ \\ [/tex]

can also be expanded as

[tex]\sf \: {(1 + x)}^{n} = C_n + C_{n - 1}x + C_{n - 2} {x}^{2} + ... + C_{1} {x}^{n - 1} + C_0{x}^{n} \\ \\ [/tex]

On multiply above two equations, we get

[tex]\sf \: {(1 + x)}^{2n} =(C_0 + C_1x + C_2 {x}^{2} + ... + C_{n - 1} {x}^{n - 1} + C_n {x}^{n})( C_n + C_{n - 1}x + C_{n - 2} {x}^{2} + ... + C_{1} {x}^{n - 1} + C_0{x}^{n}) \\ \\ [/tex]

As we have to find the value of

[tex]\boxed{ \sf{ \:\sf \: \:C_1 + C_1C_2 + ... + C_{n - 1}C_n \: }} \\ \\ [/tex]

Now, On comparing the coefficient of [tex]x^{n-1} [/tex] on both sides, we get

[tex]\sf \: \: (^{2n}C_{n - 1}) =C_0C_1 + C_1C_2 + ... + C_{n - 1}C_n \\ \\ [/tex]

[tex]\sf \: \: \dfrac{{2n}!}{(n - 1)!(2n - n + 1)!} =C_0C_1 + C_1C_2 + ... + C_{n - 1}C_n \\ \\ [/tex]

[tex]\sf \: \: \dfrac{{2n}!}{(n - 1)!(n + 1)!} =C_0C_1 + C_1C_2 + ... + C_{n - 1}C_n \\ \\ [/tex]

can be rewritten as

[tex]\sf \: \:C_1 + C_1C_2 + ... + C_{n - 1}C_n = \dfrac{{2n}!}{(n - 1)!(n + 1)!} \\ \\ [/tex]

[tex]\qquad\boxed{ \sf{ \: \because \: C_0 = 1 \: }} \\ \\ [/tex]

Hence,

[tex]\sf\implies \sf \: \:C_1 + C_1C_2 + ... + C_{n - 1}C_n = \dfrac{{2n}!}{(n - 1)!(n + 1)!} \\ \\ [/tex]