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# A man 8 friends whom he wants to invite for dinner. The number of ways

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Re: A man 8 friends whom he wants to invite for dinner. The number of ways [#permalink]
Top Contributor
Mo2men wrote:

Hi Brent,

Can you show another approach?

What 'at least' mean?

"at least" means "greater than or equal to"

So, if I say that I own at least 3 guitars, then the number of guitars I own = 3 or 4 or 5 or 6....

Cheers,
Brent
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Re: A man 8 friends whom he wants to invite for dinner. The number of ways [#permalink]
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VyshakhR1995 wrote:
A man 8 friends whom he wants to invite for dinner.The number of ways in which he can invite at least one of them is

A) 8
B) 255
C) 8!-1
D) 256
E) 7

Another approach (for Mo2men )

Number of ways to invite at least 1 friend = (# of ways to invite exactly 1 friend) + (# of ways to invite exactly 2 friends) + (# of ways to invite exactly 3 friends) + (# of ways to invite exactly 4 friends) + . . . . + (# of ways to invite exactly 8 friends)

Since the order in which we invite the friends does not matter, we can use COMBINATIONS.
For example, the number of ways to invite exactly 2 friends = 8C2
And the number of ways to invite exactly 3 friends = 8C3
etc.

So, the number of ways to invite at least 1 friend = 8C1 + 8C2 + 8C3 + . . . + 8C7 + 8C8
= 8 + 28 + 56 + 70 + 56 + 28 + 8 + 1
= 255

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Re: A man 8 friends whom he wants to invite for dinner. The number of ways [#permalink]
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VyshakhR1995 wrote:
A man 8 friends whom he wants to invite for dinner.The number of ways in which he can invite at least one of them is

A)8
B)255
C)8!-1
D)256
E)7

We can use the following equation:

Total number of ways to invite friends = number of ways to invite at least one friend - number of ways to invite zero friends.

The total number of ways the man can invite his friends is 2^8 since he can or cannot invite each friend. The number of ways to invite no friends is 1. Thus, the number of ways to invite at least one friend is 2^8 - 1 = 256 - 1 = 255.

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Re: A man 8 friends whom he wants to invite for dinner. The number of ways [#permalink]
The question could be interpreted in another way. "If we will flip a coin 8 times what is the number of ways we get at least one tail"

When you flip a coin 8 times there are 2^8 different ways the result may come out. One of the possible results is to get 8 heads - HHHHHHHH. Now you can imagine tail is to invite a friend and head is not to invite. Getting 8 heads means not to invite any of them.

Therefore, all outcomes satisfy us except when we get eight heads. There is only one way to get 8 heads. So we deduct this one outcome from total number of possible outcomes
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Re: A man 8 friends whom he wants to invite for dinner. The number of ways [#permalink]
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VyshakhR1995 wrote:
A man 8 friends whom he wants to invite for dinner. The number of ways in which he can invite at least one of them is

A)8
B)255
C)8!-1
D)256
E)7

Source :Quantpdf

An alternate approach for this question would be to find the subsets of a set with 8 entities.

Lets take a smaller set, say there were only 3 friends (A, B & C) then the ways to invite at least 1 would be A, B, C, AB, AC, BC and ABC (7 ways). This is $$2^n-1$$ $$[2^3-1 = 7]$$ in terms of the formula to find the subset of a set with n entities. We subtracted 1 because $$2^n$$ also includes an empty set, whereas we have been provided a constraint of at least 1.

Applying the same to the problem at hand, there are 8 friends, hence $$2^8-1 = 256-1 = 255$$ (Ans B)
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Re: A man 8 friends whom he wants to invite for dinner. The number of ways [#permalink]
VyshakhR1995 wrote:
A man 8 friends whom he wants to invite for dinner. The number of ways in which he can invite at least one of them is

A)8
B)255
C)8!-1
D)256
E)7

Source :Quantpdf

Given: A man 8 friends whom he wants to invite for dinner.

Asked: The number of ways in which he can invite at least one of them is

No of ways to invite at least 1 friends = $$^8C_1 + ^8C_2 + ^8C_3 + ^8C_4 + ^8C_5 + ^8C_6 + ^8C_7 + ^8C_8$$

$$(1+x)^n = ^nC_0 x^0 + ^nC_1 x + ^nC_2 x^2 + ... + ^nC_r x^r +.... + ^nC_n x^n$$

n =8
$$(1+x)^8 = ^8C_0 x^0 + ^8C_1 x + ^8C_2 x^2 + ... + ^8C_r x^r +.... + ^8C_8 x^n$$
When x =1
$$2^8 = 1 + ^8C_1 + ^8C_2 + ^8C_3 + ^8C_4 + ^8C_5 + ^8C_6 + ^8C_7 + ^8C_8$$
$$2^8 -1= ^8C_1 + ^8C_2 + ^8C_3 + ^8C_4 + ^8C_5 + ^8C_6 + ^8C_7 + ^8C_8 = 256-1 =255$$

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Re: A man 8 friends whom he wants to invite for dinner. The number of ways [#permalink]
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Re: A man 8 friends whom he wants to invite for dinner. The number of ways [#permalink]
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