Combinations problems and word problems

Question 1

Question:  

Answer:   
As n represents all positive integers, we have

Multiplying the above terms of both sides respectively, we get


Multiplying both sides of inequality by n!, we get





or n < 2n

Multiplying the above terms of both sides respectively, we get

Multiplying both sides by n!, we get


From (1) and (2), we get

Question 2

Question:  





vi) (n! + 1) is not divisible by any natural number between 2 and n.
vii) Simplify

Answer:   




















= (32.16.8.4.2)(33.31.30.18.17.15.14.13.12.11.10....3.1)
= (2⁵.2⁴.2³.2².2¹)(33.31.30.....3.1)

which is divisible by 2¹⁵.

vi) (n! + 1) is not divisible by any natural number between 2 and n.

Clearly, n! is divisible by every natural number from 2 to n.
n! + 1 will leave remainder 1, when divided by any natural number
between 2 and n.
Hence, (n! + 1) is not divisible by any natural number from 2 to n.

Question 3

Question:  

Answer:   

Question 4

Question:   Prove that n!(n + 2) = n! + (n + 1)!.

Answer:    L.H.S = n!(n + 2)
= n![(n+1)+1]

Question 5

Question:  

Answer:   

Question 6

Question:   Find n, if P(n-1,3): P(n+1,3):: 5 : 12.

Answer:   




n = 8 (as n cannot be a fraction)

Question 7

Question:   How many three letter words (with or without meaning) can be formed out of the letters of the word LOGARITHMS if repetition of letters is not allowed?

Answer:    Number of letters in the word LOGARITHMS is 10.
The three letter words formed by using these letters will be

Question 8

Question:   There are three different rings to be worn in 5 fingers with at most one in each finger. In how many ways can this be done?

Answer:    For the first ring, there are 4 options as one finger cannot have more
than one ring. The remaining two fingers will have 3 and 2 options respectively. Hence, the number of ways = 4.3.2 = 24 ways.

Question 9

Question:   How many six digits telephone numbers can be made if each number starts with 45 and no digit appears more than once?

Answer:    The first two places are reserved for 4 and 5.
The remaining 8 digits can be arranged in P(8,4) = 8.7.6.5
= 1680 ways

Question 10

Question:   Twelve students are participating in a competition. In how many ways can the first 3 prizes be won?

Answer:    Number of students participating in the competition = 12
The number of ways in which the first three prizes can be won
= P (12,3) = 12 x 11 x 10 = 1320
Hence, 12 students can win three prizes in 1320 ways.