Maths Quiz on Permutations and Combinations for JEE Mains/Advance

a) 325

Solution:

Case I: When only one flag is used. No. of signals made = ⁵P₁ = 5.

Case II: When only two flags are used. Number of signals made = ⁵P₂ = 5 * 4 = 20.

Case III: When only three flags are used. Number of signals is made = ⁵P₃ = 5 * 4 * 3 = 60.

Case IV : When only four flags are used. Number of signals made = ⁵P₄ = 5 * 4 * 3 * 2 = 120.

Case V : When five flags are used. Number of signals made = ⁵P₅ = 5! = 120.

Hence, required number = 5 + 20 + 60 + 120 + 120 = 325.

a) 601

Solution:

(1) Alphabetical order is A, C, H, I, N, S

No. of words starting with A – 5!

No. of words starting with C – 5!

No. of words starting with H – 5!

No. of words starting with I – 5!

No. of words starting with N – 5!

SACHIN-1

601.

c) 879

Solution:

p = 10, q = 9, r = 7

Total ways of selection = (p + 1) * (q + 1) * (r + 1) – 1

= [11 * 10 * 8] – 1

= 879

c) 999

Solution:

Two rings may have the same letter at a time, but the same ring cannot have two letters at a time. Therefore, we must proceed ring wise. Each of the three rings can have any one of the 10 different letters in 10 ways.

Therefore, the total number of attempts = 10 × 10 × 10 = 1000.

But out of these 1000 attempts, only one attempt is successful.

Therefore, the required number of unsuccessful attempts = 1000 – 1 = 999.

b) 30

Solution:

We have,

x* y * z = 24

x* y * z = 2³ × 3¹

The number of ways of distributing ‘n’ identical balls into ‘r’ different boxes is ^{(n + r − 1)}C_{(r − 1)}

Here we have to group 4 numbers into three groups

Number of integral positive solutions

= ^{(3 + 3 − 1)}C_{(3 − 1)} × ^{(1+ 3 − 1)}C_{(3 − 1)}

= ⁵C₂ × ³C₂

= 30