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How many different ways can $10$ people be divided into $5$ groups of two people each.
Option: 1
$945$

Option: 2
$120$

Option: 3
$252$

Option: 4
$126$

Answers (1)

best_answer

The given information is:

Number of people $=10$

Number of teams $=5$

Number of people in each team $=2$

Pick first $2$ people from $10$ people,

Total no. of ways of choosing is:

$\frac{10 !}{(10-2) ! 2 !}=\frac{10 !}{8 ! 2 !}=45$

Pick first $2$ people from $8$ people,

Total no. of ways of choosing is:

$\frac{8 !}{(8-2) ! 2 !}=\frac{8 !}{6 ! 2 !}=28$

Pick first $2$ people from $6$ people,

Total no. of ways of choosing is:

$\frac{6 !}{(6-2) ! 2 !}=\frac{6 !}{4 ! 2 !}=15$

Pick first $2$ people from $4$ people,

Total no. of ways of choosing is:

$\frac{4 !}{(4-2) ! 2 !}=\frac{4 !}{2 ! 2 !}=6$

Pick last $2$ people from $2$ people and the number of ways this can be done is $1$ .
Since the five groups are similar, there is no differentiation between them. Therefore, we need to divide it with $5 !=120$ .
The total numbers of ways:

$\frac{45 \times 28 \times 15 \times 6}{120}=\frac{113400}{120}=945$

Posted by

Shailly goel

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