Formula:- (i)nCr(i) No girlsTotal number of ways=4C0. 7C5=21(ii) at least one boy and one girlCASE(A)1 boy and 4 girls=7C1. 4C4=7CASE(B)2 boys and3 girls=7C2. 4C3=84CASE(C) 3boys and 2girls=7C3. 4C2=210CASE (D)4 boys and 1 girls=7C4. 4C1=140Total 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

Q. 255.0( 1 Vote )

A group consists

Class 11th Mathematics - Exemplar 7. Permutations and Combinations

Answer :

Formula:- (i)ⁿC_r

(i) No girls

Total number of ways

=⁴C_0. ⁷C₅

=21

(ii) at least one boy and one girl

CASE(A)1 boy and 4 girls

=⁷C_1. ⁴C₄

CASE(B)2 boys and3 girls

=⁷C_2. ⁴C₃

=84

CASE(C) 3boys and 2girls

=⁷C_3. ⁴C₂

=210

CASE (D)4 boys and 1 girls

=⁷C_4. ⁴C₁

=140

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

=7+84+210+140

=441

(iii) At least three girls

=⁴C_3. ⁷C₂+⁴C_4.⁷C₁

= 4 × 21 + 7

= 84 + 7

= 91

Rate this question :

How useful is this solution?

We strive to provide quality solutions. Please rate us to serve you better.

PREVIOUSA sports team of NEXTIf nC<

Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts

Dedicated counsellor for each student

24X7 Doubt Resolution

Daily Report Card

Detailed Performance Evaluation

view all courses