Math, asked by tejasbgowda, 11 months ago

a group consists of 7 boys and 5 girls find the number of ways in which a team of 5 members can be selected so has to have atleast one boy and one girl​

Answers

Answered by chbilalakbar
42

Answer:

"770 ways"

Step-by-step explanation:

There are 7 boys and 5 girls in a group

Total number of members  = 7 + 5 = 12

If we want to select team of 5 members from group of 12 members in such a way that there is a girl and a boys at least present in a team

Strategy :

We use the combination to solve it

Let suppose we select the team in which all the girls then

Number of teams = ⁵C₅ =  5!/(5!×0!) = 1

If we select all the boys to make team of 5 members then

Number of teams = ⁷C₅ = 7!/(5!×2!) = 21

Now

If we select the team of 5 members from 12 members then

Number of teams = ¹²C₅= 12!/(5!×7!) = 792

Number of ways at least one girl & one boys =  Total number - only boys - only girls

= 792 - 21 - 1

= 770

There are total 770 number of ways which fulfill the desire condition


amitnrw: Permutation can not be applied here as order does not matter here
chbilalakbar: Thanks Sir.You are really a good teacher of mathematics.
Answered by amitnrw
17

Answer:

770

Step-by-step explanation:

a group consists of 7 boys and 5 girls find the number of ways in which a team of 5 members can be selected so has to have atleast one boy and one girl​

Total members = 7  + 5 = 12

Total number of ways a team  of 5 can be created

= ¹²C₅

= 12!/(5!7!)

= 792

If only boys are selected then  ⁷C₅

= 7!/(5!2!)

= 21

If only girls are selected then  ⁵C₅

= 5!/(5!0!)

= 1

Number of ways at least one girl & one boys =  Total number - only boys - only girls

= 792 - 21 - 1

= 770

There are 770 number of ways in which a team of 5 members can be selected so has to have atleast one boy and one girl​


chbilalakbar: Thanks.You are really a good mathematician.
amitnrw: Welcome. Thanks for appreciation
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