Answer :

Formula:- (i)nCr


(i) No girls


Total number of ways


=4C0. 7C5


=21


(ii) at least one boy and one girl


CASE(A)1 boy and 4 girls


=7C1. 4C4


=7


CASE(B)2 boys and3 girls


=7C2. 4C3


=84


CASE(C) 3boys and 2girls


=7C3. 4C2


=210


CASE (D)4 boys and 1 girls


=7C4. 4C1


=140


Total number of ways= CASE(A)+CASE(B)+CASE(C)+CASE(D)


=7+84+210+140


=441


(iii) At least three girls


=4C3. 7C2+4C4.7C1


= 4 × 21 + 7


= 84 + 7


= 91


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