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Jane must select three different items for each dinner she

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Jane must select three different items for each dinner she [#permalink] New post 27 Sep 2017, 14:17
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Jane must select three different items for each dinner she will serve. The items are to be chosen from among five different vegetarian and four different meat selections. If at least one of the selections must be vegetarian, how many different dinners could Jane create?

A 30

B 40

C 60

D 70

E 80
[Reveal] Spoiler: OA

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Re: Jane must select three different items for each dinner she [#permalink] New post 27 Sep 2017, 21:54
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Carcass wrote:
Jane must select three different items for each dinner she will serve. The items are to be chosen from among five different vegetarian and four different meat selections. If at least one of the selections must be vegetarian, how many different dinners could Jane create?

A 30

B 40

C 60

D 70

E 80


there are many ways to solve the problem -((Atleast one veg means 1 Veg or 2 Veg or all 3 Veg dishes.))

SInce Jane always have 1 veg dish, we can take the following combination-

1veg dish AND 2 non-veg dishes = 5C1 * 4C2 = 5*6 = 30 ways

or 2veg dishes AND 1 non-veg dish = 5C2 * 4C1 = 10*4 = 40 ways

or 3 veg dishes = 5C3 = 10 ways

Total no. of ways Jane can create different dinners = 30+40+10 = 80 ways


Alternative way-

The number of ways to choose 3 items from the total of 9 items is 9C3 = 84.

The number of ways to choose only meat is 4C3 = 4.

Therefore, there are 84-4 = 80 different dinners
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Re: Jane must select three different items for each dinner she [#permalink] New post 24 Nov 2018, 07:11
I used the following approach. Can someone tell what is wrong with this approach -:

1 veg should always be there = 5 ways in which this can be selected
now we are left with 4 veg and 4 non-veg dishes
so no of ways in which 2 items can be chosen from a total of 8 = 8C2= 28
hence total no of ways = 28*5 = 140
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Re: Jane must select three different items for each dinner she [#permalink] New post 24 Nov 2018, 07:24
Expert's post
kunalkmr62 wrote:
I used the following approach. Can someone tell what is wrong with this approach -:

1 veg should always be there = 5 ways in which this can be selected
now we are left with 4 veg and 4 non-veg dishes
so no of ways in which 2 items can be chosen from a total of 8 = 8C2= 28
hence total no of ways = 28*5 = 140



you are going wrong because there are repetitions in it..
let me give you an example with a smaller set of numbers...

say 2 veg A and B AND 1 non veg C..
so atleast 1 veg should be there and we are looking for ways to choose two ..
By your method..
choose one of two say 2C1 and then 1 from remaining veg and 1 non veg so 1 out of two therefore 2C1
total 2C1*2C1=2*2=4
But is it so...NO
let us see the 4 ways ..
1) choose A, the other could be B or C.. AB and AC
2) choose B, the other could be A or C.. BA and BC
Isnt AB and BA same, so repetitions...

actual answer would be
both veg - 1
one veg and one non veg - 2C1*1=2
total 3

Hope it helps
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Some useful Theory.
1. Arithmetic and Geometric progressions : https://greprepclub.com/forum/progressions-arithmetic-geometric-and-harmonic-11574.html#p27048
2. Effect of Arithmetic Operations on fraction : https://greprepclub.com/forum/effects-of-arithmetic-operations-on-fractions-11573.html?sid=d570445335a783891cd4d48a17db9825
3. Remainders : https://greprepclub.com/forum/remainders-what-you-should-know-11524.html
4. Number properties : https://greprepclub.com/forum/number-property-all-you-require-11518.html
5. Absolute Modulus and Inequalities : https://greprepclub.com/forum/absolute-modulus-a-better-understanding-11281.html

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Re: Jane must select three different items for each dinner she [#permalink] New post 24 Nov 2018, 23:31
By my method , to chose 1 veg we have 2 ways and now we need to chose 2 items from the remaining 2 , which can be done in only 1 way or 2C2 ways.
Hence total no of dinner possible = 2 . Same as yours

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Re: Jane must select three different items for each dinner she [#permalink] New post 25 Nov 2018, 00:18
Expert's post
kunalkmr62 wrote:
By my method , to chose 1 veg we have 2 ways and now we need to chose 2 items from the remaining 2 , which can be done in only 1 way or 2C2 ways.
Hence total no of dinner possible = 2 . Same as yours

Posted from my mobile device Image


No our answers are different, my answer here would be to choose two out of 3 =3C2=3..

Anyways the reason to tell you about that was the flaw in the approach and why you were getting wrong answers.
If you have got it, it is good.
_________________

Some useful Theory.
1. Arithmetic and Geometric progressions : https://greprepclub.com/forum/progressions-arithmetic-geometric-and-harmonic-11574.html#p27048
2. Effect of Arithmetic Operations on fraction : https://greprepclub.com/forum/effects-of-arithmetic-operations-on-fractions-11573.html?sid=d570445335a783891cd4d48a17db9825
3. Remainders : https://greprepclub.com/forum/remainders-what-you-should-know-11524.html
4. Number properties : https://greprepclub.com/forum/number-property-all-you-require-11518.html
5. Absolute Modulus and Inequalities : https://greprepclub.com/forum/absolute-modulus-a-better-understanding-11281.html

Re: Jane must select three different items for each dinner she   [#permalink] 25 Nov 2018, 00:18
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Jane must select three different items for each dinner she

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