Given that,
A group consists of 4 girls and 7 boys.
We need to select a team of 5 members with the following conditions :
i. If a team has no girl
ii. If a team has at least one boy and one girl
iii. If a team has at least 3 girls.
i. Given that we need to select a team of 5 members with no girl present in it out of 4 girls and 7 boys.
Let us assume the no. of ways of selection be N
⇒ N = (selecting 5 members out of 7 boys without any girl)
⇒ N = 7C5
We know that,
nCr = \(\frac{n!}{(n-r)!r!}\)
And also,
n! = (n)(n – 1)......2.1
The no. of ways of selecting 5 members without a girl is 21 ways.
ii. Given that we need to select team of 5 members with at least 1 boy and 1 girl.
Let us assume the no. of ways of selection be N1.
⇒ N1 = (Total ways of selecting 5 members out of all 11 members) – (No. of ways of selecting 5 members without any girl)
⇒ N1 = (11C5)–(7C5)
We know that,
nCr = \(\frac{n!}{(n-r)!r!}\)
And also,
n! = (n)(n – 1)......2.1
⇒ N1 = 462–21
⇒ N1 = 441
The no. of ways of selecting 5 members with at least 1 girl and 1 boy is 441.
iii. Given that we need to find the no. of ways to select 5 members with at least 3 girls out of 7 boys and 4 girls.
The following are the possible cases :
i. Selecting 3 girls and 2 boys
ii. Selecting 4 girls and 1 boy
Let us assume the no. of ways of selection be N2.
⇒ N2 = (No. of ways of selecting 3 girls and 2 boys out of 7 boys and 4 girls) + (No. of ways of selecting 4 girls and 1 boy out of 7 boys and 4 girls)
⇒ N2 = ((4C3) × (7C2)) + ((4C4) × (7C1))
We know that,
nCr = \(\frac{n!}{(n-r)!r!}\)
And also,
n! = (n)(n – 1)......2.1
⇒ N2 = (4 × 21) + (1 × 7)
⇒ N2 = 84 + 7
⇒ N2 = 91
The no. of ways of selecting 5 members with at least 3 girls is 91.