Question

The number of ways of choosing 10 objects out of 31 objects of which 10 are identical and the remaining 21 are distinct, is :
(A) 2²⁰ (B) 2²⁰ + 1
(C) 2²¹ (D) 2²⁰ – 1

Answer 1

Step-by-step explanation:

Given The number of ways of choosing 10 objects out of 31 objects of which 10 are identical and the remaining 21 are distinct, is

• Now we have 31 objects.
• Out of which 10 are identical and the remaining 21 are different (distinct)
• Now we need to select 10
• So possible ways will be 0 identical + 10 distinct = 1 x 21 C10
•                                         1 identical + 9 distinct = 1 x 21 C9
•                                         2 identical + 8 distinct = 1 x 21 C8
•                                          3                + 7
•                                           4                + 6
•                                          ……………………….
•                                          10 identical + 0 distinct = 1 x 21C0
• Therefore total number of ways will be
•                              21 C10 + 21 C9 + 21 C8+…………= x --------(1 )
• So we know that nCr = n C n – r
•                        21 C11 + 21 C12 + 21 C13 +…………+ 21 C21 = x ------(2)
• Adding (1) and (2) we get
•                     2x = 21 C0 + 21 C1 + 21 C2 +-------------+ 21 C21
•                (So we have nC0 + nC1 +----------+nCn = 2^n)
•              So 2x = 2^21
•                    x = 2^21 / 2
•                   x = 2^21 – 1
•               Or x = 2^20
• So required number of ways will be
•                    21 C10 + 21 C9 +--------------+ 21 C0 = 2^20

Reference link will be

https://brainly.in/question/22631202

https://brainly.in/question/3295363