Combination??

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shilpaqueen: Junior | Next Rank: 30 Posts; Posts: 19; Joined: Thu Jul 29, 2010 2:12 am

by shilpaqueen » Sun Nov 21, 2010 10:22 am

Moctail has to be made from 3 kinds of wiskey, four kinds of soda and 5 kinds of juice. How many mocktails can be made using atleast 1 of each kind

135
235
325
3255
5235

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Source: Beat The GMAT — Problem Solving | Sun Nov 21, 2010 10:22 am

Geva@EconomistGMAT: GMAT Instructor; Posts: 905; Joined: Sun Sep 12, 2010 1:38 amFollowed by: 123 members; Thanked: 378 times; GMAT Score:760

by Geva@EconomistGMAT » Mon Nov 22, 2010 7:27 am

i believe the question is flawed. What does "at least one of each" mean? Can I use more than one each, e.g. 2 whiskeys, 3 sodas and 2 of the juices? If that is the case, the question has too many scenarios to be solved under GMAT conditions.

Sample scenarios:
1W out of 3, 1S out of 4, 1J out of 5: number of options = 3*4*5=60
1W out of 3, 2S out of 4, 1J out of 5: number of options = 3 * 4C2 * 5 = 3 * (4*3/2) * 5 = 3*6*5 = 90
1W out of 3, 2S out of 4, 3J out of 5: number of options = ?

etc. etc.

On the other hand, if the question meant "using 1 of each kind", the answer would be a simple 3*4*5=60 - which, unfortunately, is not an answer choice.

What's the source? OA?

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Mon Nov 22, 2010 8:06 am

A mocktail is to be prepared from 3 kinds of whiskies, 4 kinds of soda and 5 kinds of fruit juices. How many mocktails can be made that include at least 1 kind of each ingredient?

A. 135
B. 235
C. 325
D. 3255
E. 5235

I was asked by PM to comment.

Given n choices, the number of ways to select at least 1 of the n choices = 2^n - 1.

For example, given 3 people, the number of ways to choose at least 1 of the 3 people = 2^3 - 1 = 8-1 = 7.

Here all the ways to choose at least 1 of the 3 people:
3C1 = 3
3C2 = 3
3C3 = 1
3+3+1 = 7. Same answer as above.

In the problem above:
Number of ways to choose at least 1 of the 3 whiskeys = 2^3 - 1 = 7.
Number of ways to choose at least 1 of the 4 sodas = 2^4 - 1 = 15.
Number of ways to choose at least 1 of the 5 fruit juices = 2^5 - 1 = 31.

Now we multiply the number of choices we have of each ingredient:
7*15*31 = 3255.

The correct answer is D.

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goyalsau: Legendary Member; Posts: 866; Joined: Mon Aug 02, 2010 6:46 pm; Location: Gwalior, India; Thanked: 31 times

by goyalsau » Mon Nov 22, 2010 10:31 pm

GMATGuruNY wrote: Given n choices, the number of ways to select at least 1 of the n choices = 2^n - 1.

For example, given 3 people, the number of ways to choose at least 1 of the 3 people = 2^3 - 1 = 8-1 = 7.

Here all the ways to choose at least 1 of the 3 people:
3C1 = 3
3C2 = 3
3C3 = 1
3+3+1 = 7. Same answer as above.

Its very nice that we have a formula for choosing at least 1 from n

Can we proof it, Why it that so....

Saurabh Goyal
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Rahul@gurome: GMAT Instructor; Posts: 1179; Joined: Sun Apr 11, 2010 9:07 pmFollowed by: 88 members; Location: Milpitas, CA; Thanked: 447 times

by Rahul@gurome » Mon Nov 22, 2010 11:56 pm

Its very nice that we have a formula for choosing at least 1 from n

Can we proof

1 thing can be selected in nC1 ways, 2 things can be selected in nC2 ways.....n things can be selected in nCn ways.
So the number of ways of selecting at least 1 thing out of n things is nC1 + nC2 +...+nCn.
Now the binomial expansion of (1+x)^n = 1+nC1*x + nC2*x^2 + ......+nCn*x^n.
Replace x by 1 on both sides of the equation.
So we get (1 + 1)^n = 1+nC1*1 + nC2*1^2 + ......+nCn*1^n.
Or 2^n = 1+nC1*1 + nC2*1^2 + ......+nCn*1^n = 1+nC1 + nC2 +....+nCn.
So 2^n - 1 = nC1 + nC2 +...+nCn.
Hence the number of ways of selecting at least 1 thing out of n things is 2^n - 1.

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goyalsau: Legendary Member; Posts: 866; Joined: Mon Aug 02, 2010 6:46 pm; Location: Gwalior, India; Thanked: 31 times

by goyalsau » Mon Nov 22, 2010 11:59 pm

Rahul@gurome wrote:
Its very nice that we have a formula for choosing at least 1 from n

Can we proof
1 thing can be selected in nC1 ways, 2 things can be selected in nC2 ways.....n things can be selected in nCn ways.
So the number of ways of selecting at least 1 thing out of n things is nC1 + nC2 +...+nCn.
Now the binomial expansion of (1+x)^n = 1+nC1*x + nC2*x^2 + ......+nCn*x^n.
Replace x by 1 on both sides of the equation.
So we get (1 + 1)^n = 1+nC1*1 + nC2*1^2 + ......+nCn*1^n.
Or 2^n = 1+nC1*1 + nC2*1^2 + ......+nCn*1^n = 1+nC1 + nC2 +....+nCn.
So 2^n - 1 = nC1 + nC2 +...+nCn.
Hence the number of ways of selecting at least 1 thing out of n things is 2^n - 1.

How you did this i m not able to understand.

Saurabh Goyal
[email protected]
-------------------------

EveryBody Wants to Win But Nobody wants to prepare for Win.

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shovan85: Community Manager; Posts: 991; Joined: Thu Sep 23, 2010 6:19 amFollowed by: 24 members; Location: Bangalore, India; Thanked: 146 times

by shovan85 » Tue Nov 23, 2010 1:21 am

As per the concept this is what I understand:

Two ppl say A and B:

Total possible selection in 2^2 = 4 ways
This 4 ways can be A, B, AB or BA.

At least one person is selected = 2^2 - 1 = 3 ways
This 3 ways can be either A, AB,BA or B, AB, BA. (See in first set A is the at least person to selected and in second B s the at least person to selected )

goyalsau wrote:
Rahul@gurome wrote: Now the binomial expansion of (1+x)^n = 1+nC1*x + nC2*x^2 + ......+nCn*x^n.

How you did this i m not able to understand.

The Bold part is a Binomial Expansion, which describes the algebraic expansion of powers of a binomial. According to the theorem, it is possible to expand the power (x + y)^n into a sum involving terms of the form a(x^b)(y^c), where the coefficient of each term is a positive integer, and the sum of the exponents of x and y in each term is n.

The general formula is:

(1+x)^n = 1+nC1*x + nC2*x^2 + ......+nCn*x^n.

For more info: https://en.wikipedia.org/wiki/Binomial_theorem

So what we need to prove by applying this? We need to show 2^n - 1 = nC1 + nC2 + .... +nCn

2^n

= (1 + 1) ^ n

Apply Binomial Expansion (See x = 1)

1 + nC1 * 1 + nC2 * 1^2 + .... + nCn * 1^n

Thus, 2^n = 1 + nC1 * 1 + nC2 * 1^2 + .... + nCn * 1^n

=> 2^n - 1 = nC1 * 1 + nC2 * 1^2 + .... + nCn * 1^n

=> 2^n - 1 = nC1 + nC2 + .... +nCn.

If you have any doubt about Binomial Expansion then you can refer the above link. I personally feel when asked with at least one is used the remember to use the formula rather thinking about the proof.

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goyalsau: Legendary Member; Posts: 866; Joined: Mon Aug 02, 2010 6:46 pm; Location: Gwalior, India; Thanked: 31 times

by goyalsau » Tue Nov 23, 2010 1:39 am

shovan85 wrote:As per the concept this is what I understand:

Two ppl say A and B:

Total possible selection in 2^2 = 4 ways
This 4 ways can be A, B, AB or BA.

At least one person is selected = 2^2 - 1 = 3 ways
This 3 ways can be either A, AB,BA or B, AB, BA. (See in first set A is the at least person to selected and in second B s the at least person to selected )

goyalsau wrote:
Rahul@gurome wrote: Now the binomial expansion of (1+x)^n = 1+nC1*x + nC2*x^2 + ......+nCn*x^n.

How you did this i m not able to understand.
The Bold part is a Binomial Expansion, which describes the algebraic expansion of powers of a binomial. According to the theorem, it is possible to expand the power (x + y)^n into a sum involving terms of the form a(x^b)(y^c), where the coefficient of each term is a positive integer, and the sum of the exponents of x and y in each term is n.

The general formula is:

(1+x)^n = 1+nC1*x + nC2*x^2 + ......+nCn*x^n.

For more info: https://en.wikipedia.org/wiki/Binomial_theorem

So what we need to prove by applying this? We need to show 2^n - 1 = nC1 + nC2 + .... +nCn

2^n

= (1 + 1) ^ n

Apply Binomial Expansion (See x = 1)

1 + nC1 * 1 + nC2 * 1^2 + .... + nCn * 1^n

Thus, 2^n = 1 + nC1 * 1 + nC2 * 1^2 + .... + nCn * 1^n

=> 2^n - 1 = nC1 * 1 + nC2 * 1^2 + .... + nCn * 1^n

=> 2^n - 1 = nC1 + nC2 + .... +nCn.

If you have any doubt about Binomial Expansion then you can refer the above link. I personally feel when asked with at least one is used the remember to use the formula rather thinking about the proof.

Thanks Buddy i asked for the Proof Because Its very Hard for me to remember the Formula , at times when I saw the same problem again I can always recall the basic reasoning behind the formula so its easy for me to recall formulas...

Saurabh Goyal
[email protected]
-------------------------

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Bek: Senior | Next Rank: 100 Posts; Posts: 68; Joined: Fri Dec 03, 2010 10:29 pmFollowed by: 1 member; Thanked: 5 times

by Bek » Fri Dec 03, 2010 10:32 pm

MaxGMAT

A mocktail has to be prepared from three kinds of whiskey, 4 kinds of soda and five kinds of fruit juice. How many mocktails can be made taking at least one of each kind?

(A) 235 (B) 325 (C) 3255 (D) 5235 (E) 565

OA is C

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