Archive for the ‘General Questions’ Category

Problem 16

Problem – 16

What is the probability that a fair die never comes up an even number when it is rolled ten times?

 

Solution :

Probability to rolled the die for one time and comes up an even number =  \frac{3}{6}= \frac{1}{2}

As the rolls are independent so probability after rolling ten times = \left ( \frac{1}{2} \right )^{10}= \frac{1}{1024}

Problem 15

Problem – 15

What is the probability that a five-card poker hand contains the three of hearts, the two of clubs and the jack of diamonds?

 

Solution :

For first three cards (three of hearts, the two of clubs and the jack of diamonds ) we do have choice, so we have to choose remaining 2 cards from the remaining 49 cards. So ways to choose = C(49,2) =1176

And total number of ways to choose five cards = C(52,5) = 2598960

So probability = \frac{1176}{2598960}= \frac{1}{2210}

Problem 14

Problem – 14

How many different strings can be made from the letters in EVERGREEN, using all the letters?

 

Solution :

Total number of different string we can made from the letters EVERGREEN =  \frac{9!}{4!2!}=\frac{362880}{24\cdot 2}= 7560  

Problem 13

Problem – 13

How many ways are there to distribute seven indistinguishable balls into seven distinguishable bins?

 

Solution :

Distributing k indistinguishable balls into n distinguishable boxes, without exclusion, corresponds to forming a combination of size k with unrestricted repetitions, taken from a set of size n. Therefore, there are C(n+k-1, k) different ways to k distribute k indistinguishable balls into n distinguishable boxes, without exclusion.

Using the above theorem total number of ways = C(7+7-1,7) = C(13,7) = 1716

Problem 12

Problem – 12

How many solutions are there to the equation?

x1+x2+x3+x4+x5=23,

where xi, i=1, 2, 3, 4, 5, is a nonnegative integer such that

a) x1≥1?

b) xi≥1 for i=1, 2, 3, 4, 5?

c) 0≤x1≤7?

d) 0≤x1≤3, 1≤x2<4, and x3≥15?

 

Solution :

a)      We require x1 ≥ 1. This uses up 1 of the 23 total required, so the problem is the same as fining the number of solutions to x1’+x2’+x3’+x4’+x5’ = 22 with each xi’ a nonnegative integer. The number of solutions are therefore

C(5+22-1,22) = C(26,22) = 14950

b)      We require each xi’ ≥ 1. Thus up 5 of the 23 total required, so the problem is the same as finding the number of solutions to  x1’+x2’+x3’+x4’+x5’ = 18 with each xi’ a nonnegative integer. The number of solutions are therefore

C(5+18-1,18) = C(22,18) = 7315

c)       The number of solutions without restriction is C(5+23-1,23) = C(27,23) = 17550. The number of solutions violating the restriction by having x1≥8 is C(5+15-1,15) = C(19,15) = 3876. Therefore the answer is 17550 – 3876 = 13674

d)      The number of solutions with x2 ≥ 1 and x3 ≥ 15 ( as required) but without the restriction x1≤3 and x2<4 is C(5+7-1,7) = C(11,7) = 330. The number of solutions violating the restriction for x1, which is ≥ 0, by having x1≥ 4 is C(5+3-1,3) = C(7,3) = 35, the number of solutions violating the restriction for x2 , which is ≥ 1, by having x2≥4 is C(5+4-1,4) = C(8,4) = 70. Therefore the answer is 330 – 35 – 70 = 225

Problem 11

Problem – 11

How many ways are there to assign three jobs to nineteen employees if each employee can be given more than one job?

 

Solution:

Different ways to assign three jobs to nineteen employees = 193 = 6859

Problem 10

Problem – 10

In how many different ways can nine elements be selected in order from a set with four elements when repetition is allowed?

 

Solution:

Different ways to select 9 elements from a set of four elements = 49 = 262144

Problem 9

Problem – 9

What is the coefficient of x^{10} in (3+x)^{15}?

 

Solution:

Binomial theorem for (x+y)n = nC0xny0 + nC1xn-1y1+ nC2xn-2y2 + nC3xn-3y3 +…………+ nCn-1x1yn-1+ nCnx0yn

Using the Binomial theorem, coefficient of x10 = 15C10(3)5=3003×243 =729729

Problem 8

Problem – 8

Find the coefficient of x^8y^5 in (x+y)^{13}.

 

Solution:

Binomial theorem for (x+y)n = nC0xny0 + nC1xn-1y1+ nC2xn-2y2 + nC3xn-3y3 +…………+ nCn-1x1yn-1+ nCnx0yn

Using the Binomial theorem, coefficient of x8y5 = 13C(13-5) = 13C8 = 1287

 

Problem 7

Problem – 7

Nine women and five men are on the faculty in the mathematics department at a school.

a) How many ways are there to select a committee of four members of the department if at least one woman must be on the committee?

b) How many ways are there to select a committee of four members of the department if at least one woman and at least one man must be on the committee?

 

Solution:

 

a)      Number of ways to choose the four members if at least one woman must be on the committee = 9C1x5C3 + 9C2x5C2 + 9C3x5C1 + 9C4x5C0 = 90 + 360 + 420 + 126 = 996 ways

 

b)      Number of ways to choose the four members if at least one woman  and at least one man must be on the committee = 9C1x5C3 + 9C2x5C2 + 9C3x5C1 = 90 + 360 + 420 = 870 ways