How to do 191st question?

Question

How to do 191st question?

2 Answers

Impact of this question

1517 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-16T11:13:12

See the answer below...

Explanation:

When there is one lady, there is five gentleman in the committee. The ways to form the committee is $(^{6} C_{5} \times^{5} C_{1}) = 30$ $color(red)((""^6C_5xx""^5C_1)=30$ .

When there is two lady, there is four gentleman in the committee. The ways to form the committee is $(^{6} C_{4} \times^{5} C_{2}) = 150$ $color(red)((""^6C_4xx""^5C_2)= 150$ .

When there is three lady, there is three gentleman in the committee. The ways to form the committee is $(^{6} C_{3} \times^{5} C_{3}) = 200$ $color(red)((""^6C_3xx""^5C_3)=200$ .

When there is four lady, there is two gentleman in the committee. The ways to form the committee is $(^{6} C_{2} \times^{5} C_{4}) = 75$ $color(red)((""^6C_2xx""^5C_4)=75$ .

When there is five lady, there is one gentleman in the committee. The ways to form the committee is $(^{6} C_{1} \times^{5} C_{5}) = 6$ $color(red)((""^6C_1xx""^5C_5)=6$ .

The sum of the ways is $30 + 150 + 200 + 75 + 6 = 461$ $color(blue)(30+150+200+75+6)=461$

That is my answer, I don't know why the option doesn't match.

Answer 2 · 2018-03-16T12:02:28

$461$ $461$

Explanation:

$6$ $6$ gentlemen, $5$ $5$ ladies form $6$ $6$ -person committee, which at least

Total number of possible outcomes,

$^{11} C_{6}$ $"^11C_6$

Total number of unconditioned outcomes ( $6$ $6$ gentlemen, $0$ $0$ lady),

$^{6} C_{6} \cdot^{5} C_{0}$ $""^6C_6 *"^5C_0$

Total number of conditioned outcomes,

$^{11} C_{6} - (^{6} C_{6} \cdot^{5} C_{0})$ $""^11C_6-(""^6C_6 *"^5C_0)$

Key in your calculator,

$461$ $461$

Science

Math

Humanities

... and beyond

How to do 191st question?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question