GMAT Tough Problem Solving - 10( Permutation and Combinations , Number Series)

46.A bar is creating a new signature drink. There are five possible alcoholic ingredients in the drink: rum, vodka, gin, peach schnapps, or whiskey. There are five possible non-alcoholic ingredients: cranberry juice, orange juice, pineapple juice, limejuice, or lemon juice. If the bar uses two alcoholic ingredients and two non-alcoholic ingredients, how many different drinks are possible?


A. 100

B. 25

C. 50

D. 75

E. 3600

47. The sum of the even numbers between 1 and n is 79*80, where n is an odd number. N=?

 

48. A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together?


(A) 3510

(B) 2620

(C) 1404

(D) 700

(E) 635

49. In how many ways can the letters of the word 'MISSISIPPI' be arranged?


a) 1260

b) 12000

c) 12600

d) 14800

e) 26800

50. Goldenrod and No Hope are in a horse race with 6 contestants. How many different arrangements of finishes are there if No Hope always finishes before Goldenrod and if all of the horses finish the race?

(A) 720

(B) 360

(C) 120

(D) 24

(E) 21

Solutions

46.Soln: The first step in this problem is to calculate the number of ways of selecting two alcoholic and two non-alcoholic ingredients. Since order of arrangement does not matter, this is a combination problem.


The number of combinations of n objects taken r at a time is C(n,r) = n!/(r!(n-r!))

The number of combinations of alcoholic ingredients is

C(5,2) = 5!/(2!(3!))

C(5,2) = 120/(2(6))

C(5,2) = 10

The number of combinations of non-alcoholic ingredients is

C(5,2) = 5!/(2!(3!))

C(5,2) = 120/(2(6))

C(5,2) = 10


The number of ways these ingredients can be combined into a drink can be determined by the Multiplication Principle. The Multiplication Principle tells us that the number of ways independent events can occur together can be determined by multiplying together the number of possible outcomes for each event.


The number of possible drinks is


= 10 * 10

= 100

47.Soln: The sum of numbers between 1 and n is = (n(n+1))/2

1+2+3+.....+n=(n(n+1))/2 {formula}

we are looking for the sum of the even numbers between 1 and n, which means:

2+4+6+.....+(n-1) n is ODD

=1*2+2*2+2*3+......+2*((n-1)/2)

=2*(1+2+3+.....+*((n-1)/2))

from the formula we obtain :

=2*(((n-1)/2)*((n-1)/2+1))/2

=((n-1)/2)*((n+1)/2) =79*80

=> (n-1)*(n+1)=158*160

=> n=159

48.Soln: 4W 2M == 5C4.(8C2-1) = 5.(27) = 135

3W 3M == 5C3.(8C3-6) = 10.50 = 500

total 635


Max. number of possibilities considering we can choose any man 8c2 * 5C4 + 8C3*5c3 = 700.

consider it this way.... from my previous reply max possible ways considering we can chose any man = 700

now we know that 2 man could not be together... now think opposite... how many ways are possible to have these two man always chosen together...

since they are always chosen together...


For chosing 2 men and 4 women for the committee there is only 1 way of chosing 2 men for the committee since we know only two specific have to be chosen and


there are 5C4 ways of choosing women


1*5C4 = 5


For choosing 3 men and 3 women for the committee there are exactly 6C1 ways choosing 3 men for the committee since we know two specific have to be chosen so from the remaining 6 men we have to chose 1 and there are 5C4 ways of choosing women


6C1*5C3 = 60


So total number of unfavorable cases = 5 + 60 = 65


Now since we want to exclude these 65 cases... final answer is 700-65 = 635

49.Soln: Total # of alphabets = 10

so ways to arrange them = 10!

Then there will be duplicates because 1st S is no different than 2nd S.

we have 4 Is 3 S and 2 Ps


Hence # of arrangements = 10!/4!*3!*2!

50.Soln: two horses A and B, in a race of 6 horses... A has to finish before B


if A finishes 1... B could be in any of other 5 positions in 5 ways and other horses finish in 4! Ways, so total ways 5*4!

if A finishes 2... B could be in any of the last 4 positions in 4 ways. But the other positions could be filled in 4! ways, so the total ways 4*4!

if A finishes 3rd... B could be in any of last 3 positions in 3 ways, but the other positions could be filled in 4! ways, so total ways 3*4!

if A finishes 4th... B could be in any of last 2 positions in 2 ways, but the other positions could be filled in 4! ways, so total ways... 2 * 4!

if A finishes 5th .. B has to be 6th and the top 4 positions could be filled in 4! ways..

A cannot finish 6th, since he has to be ahead of B

therefore total number of ways

5*4! + 4*4! + 3*4! + 2*4! + 4! = 120 + 96 + 72 + 48 + 24 = 360


 

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