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2 years ago

Choose the correct formula for combination. A . n!/r!(n-r)! B. n!/(n-r)!

Mathematics

1 answer:

OleMash [197]2 years ago

4 0

Answer:

Step-by-step explanation:

The number of combinations of n objects taken r at a time is determined by the following formula: C(n,r)=n!/(n−r)!

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Please help ASAP :) this is really easy ino lol

Zanzabum

Known :

Ratio = Greg : Harry = 1 : 3

Greg's money = £8

Asked :

Harry's money = ...?

Answer :

$\: \ \: \: \: \: \: \: \frac{1}{3} = \frac{greg}{harry} \\ \: \: \: \: \: \: \: \: \frac{1}{3} = \frac{8}{harry} \\ harry = \frac{3 \times 8}{1} \\ harry = 24$

So, Harry gets £24

Hope it helpful and useful :)

5 0

3 years ago

You own 6 hats and are taking 4 on vacation. In how many ways can you choose 4 hats from the 6?

Gemiola [76]

Your answer is A. 15

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2 years ago

Read 2 more answers

Check whether the relation R on the set S = {1, 2, 3} is an equivalent

kozerog [31]

Answer:

$R$ isn't an equivalence relation. It is reflexive but neither symmetric nor transitive.

Step-by-step explanation:

Let $S$ denote a set of elements. $S \times S$ would denote the set of all ordered pairs of elements of $S\!$ .

For example, with $S = \lbrace 1,\, 2,\, 3 \rbrace$ , $(3,\, 2)$ and $(2,\, 3)$ are both members of $S \times S$ . However, $(3,\, 2) \ne (2,\, 3)$ because the pairs are ordered.

A relation $R$ on $S\!$ is a subset of $S \times S$ . For any two elements $a,\, b \in S$ , $a \sim b$ if and only if the ordered pair $(a,\, b)$ is in $R\!$ .

A relation $R$ on set $S$ is an equivalence relation if it satisfies the following:

Reflexivity: for any $a \in S$ , the relation $R$ needs to ensure that $a \sim a$ (that is: $(a,\, a) \in R$ .)

Symmetry: for any $a,\, b \in S$ , $a \sim b$ if and only if $b \sim a$ . In other words, either both $(a,\, b)$ and $(b,\, a)$ are in $R$ , or neither is in $R\!$ .

Transitivity: for any $a,\, b,\, c \in S$ , if $a \sim b$ and $b \sim c$ , then $a \sim c$ . In other words, if $(a,\, b)$ and $(b,\, c)$ are both in $R$ , then $(a,\, c)$ also needs to be in $R\!$ .

The relation $R$ (on $S = \lbrace 1,\, 2,\, 3 \rbrace$ ) in this question is indeed reflexive. $(1,\, 1)$ , $(2,\, 2)$ , and $(3,\, 3)$ (one pair for each element of $S$ ) are all elements of $R\!$ .

$R$ isn't symmetric. $(2,\, 3) \in R$ but $(3,\, 2) \not \in R$ (the pairs in $\! R$ are all ordered.) In other words, $3$ isn't equivalent to $2$ under $R\!$ even though $2 \sim 3$ .

Neither is $R$ transitive. $(3,\, 1) \in R$ and $(1,\, 2) \in R$ . However, $(3,\, 2) \not \in R$ . In other words, under relation $R\!$ , $3 \sim 1$ and $1 \sim 2$ does not imply $3 \sim 2$ .