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aleksandr82 [10.1K]
3 years ago
6

Simplify the following expressing. Show your work and explain each step. 7-(5-3(2+6(2^2)))

Mathematics
1 answer:
Anon25 [30]3 years ago
5 0

Answer:

80

Step-by-step explanation:

7-(5-3(2+6(2^2)))

When a question has multiple signs, the rule of BODMAS is used. That is, Bracket Of Division Multiplication Addition and Subtraction. The question above will therefore be solved in that order.

The first bracket, we have 2^2

That gives 4

Rewriting the question will give

7-(5-3(2+6(4)))

The question in the next bracket will follow

(2+6(4))

In this bracket too, we have the plus sign and the multiplication sign, so the multiplication will first be solved, them addition will follow.

(2+24)

(26)

Rewriting the question again, we have

7-(5-3(26))

And then, we have the last bracket

(5-3(26))

But we have both the Subtraction sign and the Multiplication sign in this bracket also, so the multiplication will first be solved, before the Subtraction.

(5-3(26))

(5-78)

(-73)

And lastly we have

7-(-73)

7+73

80

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Answer:

a) 182 possible ways.

b) 5148 possible ways.

c) 1378 possible ways.

d) 2899 possible ways.

Step-by-step explanation:

The order in which the cards are chosen is not important, which means that we use the combinations formula to solve this question.

Combinations formula:

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

In this question, we have that:

There are 52 total cards, of which:

13 are spades.

13 are diamonds.

13 are hearts.

13 are clubs.

(a)Two-pairs: Two pairs plus another card of a different value, for example:

2 pairs of 2 from sets os 13.

1 other card, from a set of 26(whichever two cards were not chosen above). So

T = 2C_{13,2} + C_{26,1} = 2*\frac{13!}{2!11!} + \frac{26!}{1!25!} = 182

So 182 possible ways.

(b)Flush: five cards of the same suit but different values, for example:

4 combinations of 5 from a set of 13(can be all spades, all diamonds, and hearts or all clubs). So

T = 4*C_{13,5} = 4*\frac{13!}{5!8!} = 5148

So 5148 possible ways.

(c)Full house: A three of a kind and a pair, for example:

4 combinations of 3 from a set of 13(three of a kind ,c an be all possible kinds).

3 combinations of 2 from a set of 13(the pair, cant be the kind chosen for the trio, so 3 combinations). So

T = 4*C_{13,3} + 3*C_{13.2} = 4*\frac{13!}{3!10!} + 3*\frac{13!}{2!11!} = 1378

So 1378 possible ways.

(d)Four of a kind: Four cards of the same value, for example:

4 combinations of 4 from a set of 13(four of a kind, can be all spades, all diamonds, and hearts or all clubs).

1 from the remaining 39(do not involve the kind chosen above). So

T = 4*C_{13,4} + C_{39,1} = 4*\frac{13!}{4!9!} + \frac{39!}{1!38!} = 2899

So 2899 possible ways.

4 0
3 years ago
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