**Method 1**:

First male can be selected in 7C1 ways.

Now, 6 men are left

First female can be selected in 6C1 ways (as one of them would be the spouse of the first male)

Now, 6 women are left

Second male can be selected in 5C1 ways (as one of the 6 males would be the spouse of first female)

Now, 5 men are left

Second female can be selected in 4C1 ways (as one of the 5 females would be the spouse of second male)

Now, 5 women are left

As when we select M1,M2 we also select M2,M1,

So, half the selections are redundant

Desired answer = 7C1*6C1*5C1*4C1/2 = **420 **

**Method 2**:

1) select unique groups of 4 people (this would be done using combinations)

2) multiply the resultant with 2 (i.e for every unique combination of 4 people there are 2 tems possible)

i.e 7C2 * 5C2 =210

i.e there are 210 unique combinations of the 4 people

so nos of teams = 2*210

=**420 **

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