Question

6) 17 students are present in a class. In how
many ways, can they be made to stand in 2
circles of 8 and 9 students?
- Published on 11 Apr 17
a. 17C, X 9! X 8!
b. 17C9 x 8! X 7!
c. 8! X 7!
d. 17Cg x 8! X 9! ​

Answer 1

Answer:

b) ¹⁷C₉ × 8! × 7!

Step-by-step explanation:

Concept:

• Combination =  $^{n} C_{r} = \frac{n!}{r!(n - r)!}$
• Combination and Arrange = Permutation = $^{2}P_{r} = \frac{n!}{(n - r)!}$

Solution:

Number of students = 17 students

First selecting or combining 9 students in circle = $^{17} C_{9}$

Arranging 9 students in circle =(9-1)! = 8! ways

The leftout students = 17 - 9 = 8

Arranging the remaining 8 students in another circle =(8-1)! = 7! ways

∴ Total ways to arrange 17 students in two circles of 8 and 9 students =

      ¹⁷C₉ × 8! × 7!

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Answer 2

Answer:

The possible number of ways = $17C9\times 8!\times7!$ and option (B) is correct.

Step-by-step explanation:

As per the data given in the question,

We have,

No. of ways of selecting 9 students from a group of 17 = 17C9

No. of ways of arranging 9 students = (n-1)! = (9-1)!= 8!

No. of ways of arranging 8 students = (n-1)! = (8-1)!= 7!

So,

The possible number of ways = $17C9\times 8!\times7!$ and option (B) is correct.

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