Problem Solving

Here I will post some interesting arrangements problems. The answers will be given later

1. There are six cars, two red, one yellow, one blue, one white, and one black. The cars are lined up in a row and cars of the same color have no difference. How many different arrangements are there so that two red cars are NOT next to each other?
(A) 720
(B) 600
(C) 480
(D) 240
(E) 120

2. Seven students, including Tom and Jerry, are to be selected a 3-people committee. If Tom and Jerry cannot be selected at the same time, in how many ways can the committee be selected?
(A) 20
(B) 28
(C) 30
(D) 35
(E) 40


3. How many free-digit numerals begin with the digit that represents a prime number and end with the digit that represents a prime number?
(A) 16
(B) 80
(C) 160
(D) 180
(E) 240


4. In how many ways can four boys and four girls be assigned seats in a row of eight seats if boys and girls are to alternate?
(A) 2!*4!*4!
(B) 4!*4!
(C) 2*4!
(D) 8!/2
(E) 8!

5. 123. How many 2-digit positive integers are there in which the sum of two digits is a 2-digit prime number?
(A) 16
(B) 17
(C) 18
(D) 19
(E) 20

Answer 1:
the six cars can be arranged in 6!/2!= 360 ways
when the two red cars are together, the cars can be arranged in 5! ways=120 ways

so when two red cars are NOT next to each other=360-120=240 ways

Answer 2:
selecting 3 out of 7=7C3 ways=35 ways
selecting Tom and Jerry together=5C1*2C2= 5 ways
so when Tom and Jerry are not together=35-5=30 ways

Answer C

Answer 3:
three digit number,
form is XYZ where X and Z must be one of 2,3,5,7
Y can be 0,1,...9
total numbers=4*10*4
=160
Answer C

Answer 4:
if boys take the odd places
total number of ways=4!*4!
same when boys take even places

total=2*4!*4!
Answer A

Answer 5:
max sum is when number is 99=18
so sum must be <18
so the prime numbers are 11,13,17

for sum to be 11, numbers are-
29
38
47
56

for sum to be 13
49
58
67
and for sum to be 17
89
total numbers=8*2(because the reverse of those numbers must be counted too)
=16
Answer A

great answers!

Can you please elaborate on Q1? Why is it 6!/2!? Is that 6P2 ?

chiller wrote:Can you please elaborate on Q1? Why is it 6!/2!? Is that 6P2 ?

its the formula:
since the two red cars are similar, there will be only one arrangement for them instead of 2.
the formula is:
if n elements and out of them r are similar, they can be arranged in n!/r! ways

Thanks for the answers tohellandback,

Can you please explain how you got 4!*4! in Qstn 4.
Is there a formula for that kind of arrangement.

Can you also tell me a place from where I can get Formula and lessons for permutation & Combinations. I am really bad at it...

Thanks

mbadreams wrote:Thanks for the answers tohellandback,

Can you please explain how you got 4!*4! in Qstn 4.
Is there a formula for that kind of arrangement.

Can you also tell me a place from where I can get Formula and lessons for permutation & Combinations. I am really bad at it...

Thanks

if the four boys are A B C D
they can be arranged in 4! ways
if the girls are M L N O
they can be arranged in 4! ways.
total arrangements= 4!*4!
this can happen in two ways because boys or the girls can take even or odd places.
total= 2*4!*4! ways

the formula just in case you don't know:
n different things can be arranged in n! ways
when two events a and b are happening, together they happen in a * b ways

mbadreams wrote:Thanks for the answers tohellandback,

Can you please explain how you got 4!*4! in Qstn 4.
Is there a formula for that kind of arrangement.

Can you also tell me a place from where I can get Formula and lessons for permutation & Combinations. I am really bad at it...

Thanks

Use the following Link! It's very useful!
https://tutors4you.com/permutationcombin ... torial.htm

Thanks a lot!!!