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Home » From 4 officers and 8 clerks, in how many ways can 6 be chosen (i) to include exactly one officer, (ii) to include at least one officer?
From 4 officers and 8 clerks, in how many ways can 6 be chosen
(i) to include exactly one officer,
(ii) to include at least one officer?
Rs aggarwal Maths CBSE Class 11 Chapter 9 Exercise 9B
From 4 officers and 8 clerks, in how many ways can 6 be chosen
(i) to include exactly one officer,
(ii) to include at least one officer?

Answer : The team of 6 has to be chosen from 4 officers and 8 clerks. There are some restrictions which are

  1. To include exactly one officer In this case ,

One officer will be chosen from 4 in 4C1 ways Therefore, 5 will be chosen form 8 clerks in 8C5 ways. Thus by multiplication principle , we get

Total no. of ways in 1 case is 4C1      8C5.

  1. To include at least one officer

In this case, there will be subcases for selection which is as follows.

  • One officer and 5 clerks
  • Two officers and 4 clerks
  • Three officers and 3 clerks
  • Four officers and 2 clerks Or

The required case of at least on officer would be

= Total cases – cases having only clerks Now,

The total case would be choosing 6 out of 12 in 12C6 ways.

And cases that would have only clerks would be i.e. selecting 6 from 8 clerks in 8C6 ways.

12C6 – 8C6 ways.

⇒924 – 28 ways

= 896 ways

i

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