A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has:

(i) No girls.

(ii) At least one boy and one girl.

(iii) At least three girls.

2 Views|Posted 7 months ago

Asked by Shiksha User

1 Answer

Sort by:

Answered by

Payal Gupta | Contributor-Level 10

7 months ago

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} W e h a v e 4 g i r l s a n d 7 b o y s a n d a t e a m o f 5 m e m b e r s i s t o b e s e l e c t e d . \\ (i) I f n o g i r l i s s e l e c t e d, t h e n a l l t h e 5 m e m b e r s a r e t o b e s e l e c t e d \\ o u t o f 7 b o y s i . e . {}_{7}{C_{5}} = \frac{7!}{5! 2!} = \frac{7 * 6.5!}{5! * 2} = 21 w a y s \\ (i i) W h e n a t l e a s t o n e b o y a n d o n e g i r l a r e t o b e s e l e c t e d, t h e n \\ N u m b e r o f w a y s = {}_{4}{C_{1}} * {}_{7}{C_{4}} + {}_{4}{C_{2}} * {}_{7}{C_{3}} + {}_{4}{C_{3}} * {}_{7}{C_{2}} + {}_{4}{C_{4}} * {}_{7}{C_{1}} \\ = 4 * \frac{7 * 6 * 5 * 4}{4 * 3 * 2 * 1} + \frac{4 * 3}{2 * 1} * \frac{7 * 6 * 5}{3 * 2 * 1} + 4 * \frac{7 * 6}{2 * 1} + 1 * 7 \\ = 4 * 35 + 6 * 35 + 4 * 21 + 7 = 140 + 210 + 84 + 7 = 441 w a y s \\ (i i i) W h e n a t l e a s t 3 g i r l s a r e i n c l u d e d, t h e n \\ N u m b e r o f w a y s = {}_{4}{C_{3}} * {}_{7}{C_{2}} + {}_{4}{C_{4}} * {}_{7}{C_{1}} \\ = 4 * \frac{7 * 6}{2 * 1} + 1 * 7 = 84 + 7 = 91 w a y s \\ H e n c e, t h e r e q u i r e d n u m b e r o f w a y s a r e \\ (i) 21 w a y s (i i) 441 w a y s (i i i) 91 w a y s \end{array}$

Upvote

Similar Questions for you

If |z + 1| = αz + β (i + 1) and  $z = \frac{1}{2}$ – 2i, find α + β.

Alok Kumar Singh

$| \frac{1}{2} - 2 i + 1 | = α (\frac{1}{2} - 2 i) + β (1 + i)$

$\sqrt[]{\frac{9}{4} + 4} = α (\frac{1}{2} - 2 i) + β (1 + i)$

$\frac{5}{2} = α (\frac{1}{2}) + β + i (- 2 α + β)$

$\frac{α}{2} + β = \frac{5}{2}$ ...(1)

–2α + β = 0 …(2)

Solving (1) and (2)

$\frac{α}{2} + 2 α = \frac{5}{2}$

$\frac{5}{2} α = \frac{5}{2}$

a =

Alok Kumar Singh

Variance = $\frac{\sum x^{2}}{n} - {(\bar{x})}^{2}$

$\frac{6 0^{2} + 6 0^{2} + 4 4^{2} + 5 8^{2} + 6 8^{2} + α^{2} + β^{2} + 5 6^{2}}{8} = {(58)}^{2} = 66.2$

$\frac{7200 + 1936 + 3364 + 4624 + 3136 + α^{2} + β^{2}}{8} = 3364 = 66.2$

$2532.5 + \frac{α^{2} + β^{2}}{8} - 3364 = 66.2$

α² + β² = 897.7 × 8

= 7181.6

Alok Kumar Singh

Start with

(1) $\bar{E} : \frac{6!}{2!} = 360$

(2) $\bar{G E} : \frac{5!}{2!},$ $\bar{G N} : \frac{5!}{2!}$

(3) GTE : 4!, GTN: 4!, GTT : 4!

(4) GTWENTY = 1

⇒ 360 + 60 + 60 + 24 + 24 + 24 + 1 = 553

Alok Kumar Singh

${(1 + x)}^{11} =^{11} C_{0} +^{11} C_{1} x +^{11} C_{2} x^{2} + . . . . . +^{11} C_{11} x^{11}$

= \frac{2^{12} - 2 - 24}{12}

= \frac{2^{12} - 26}{12} = \frac{4070}{12} = \frac{2035}{6} = \frac{m}{n}

m + n = 2035 + 6 = 2041

Let $f (x) = {\begin{matrix} 2 + 2 x, & x \in (- 1, 0) \\ 1 - \frac{x}{3}, & x \in [0, 3) \end{matrix}$ 

$g (x) = {\begin{matrix} x, & x \in [0, 1) \\ - x, & x \in (- 3, 0) \end{matrix}$ 

The range of fog(x) is

Alok Kumar Singh

$f (x) = {\begin{matrix} 2 + 2 x, & x \in (- 1, 0) \\ 1 - \frac{x}{3}, & x \in [0, 3) \end{matrix}$

$g (x) = {\begin{matrix} x, & x \in [0, 1) \\ - x, & x \in (- 3, 0) \end{matrix}$ ->g(x) = |x|, x Î (–3, 1)

$f (g (x)) = {\begin{matrix} 2 + 2 | x |, & | x | \in (- 1, 0) \Rightarrow x \in ? \\ 1 - \frac{| x |}{3}, & | x | \in [0, 3) \Rightarrow x \in (- 3, 1) \end{matrix}$

$f (g (x)) = {\begin{matrix} 1 - \frac{x}{3}, & x \in [0, 1) \\ 1 + \frac{x}{3}, & x \in (- 3, 0) \end{matrix}$

Range of fog(x) is [0, 1]

Taking an Exam? Selecting a College?

Get authentic answers from experts, students and alumni that you won't find anywhere else.

On Shiksha, get access to

66K

Colleges

1.2K

Exams

6.9L

Reviews

1.8M

Answers

Learn more about...

Maths NCERT Exemplar Solutions Class 11th Chapter Seven 2025

View Exam Details

Share Your College Life Experience

Didn't find the answer you were looking for?

Search from Shiksha's 1 lakh+ Topics

Ask Current Students, Alumni & our Experts

Quick Actions

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Have a question related to your career & education?

See what others like you are asking & answering

A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has: (i) No girls. (ii) At least one boy and one girl. (iii) At least three girls.

Similar Questions for you

Taking an Exam? Selecting a College?

Learn more about...

Maths NCERT Exemplar Solutions Class 11th Chapter Seven 2025

Share Your College Life Experience

Didn't find the answer you were looking for?

Ask Current Students, Alumni & our Experts

Quick Actions

Have a question related to your career & education?

A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has:

(i) No girls.

(ii) At least one boy and one girl.

(iii) At least three girls.