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Maths-

General

Easy

Question

From a council of 7 Hindus, 5 Muslims and 4 Christians a committee of 5 is to be formed with 1 Christian and at least two Hindus The number of ways of forming the committee is

1220
1240
1680
1280

The correct answer is: 1680

$blank to the power of 4 C subscript 1 times to the power of 7 C subscript 2 times to the power of 5 C subscript 2 plus to the power of 4 C subscript 1 times to the power of 7 C subscript 3 times to the power of 5 C subscript 1 plus to the power of 4 C subscript 1 times to the power of 7 C subscript 4 times to the power of 5 C subscript 0$

Related Questions to study

$blank to the power of straight n straight C subscript straight r plus 1 end subscript plus 2 to the power of straight n straight C subscript straight r plus to the power of straight n straight C subscript straight r minus 1 end subscript equals$

$blank to the power of straight n straight C subscript straight r plus 1 end subscript plus 2 to the power of straight n straight C subscript straight r plus to the power of straight n straight C subscript straight r minus 1 end subscript equals$

A person wishes to make up as many different parties of 10 as he can out of 20 friends, each party consisting of the same number In how many of the parties the same man is found ?

A person wishes to make up as many different parties of 10 as he can out of 20 friends, each party consisting of the same number In how many of the parties the same man is found ?

A father with 6 children takes 3 at a time to a park without taking the same children How often each child goes to the park?

A father with 6 children takes 3 at a time to a park without taking the same children How often each child goes to the park?

parallel

Action of $H N O subscript 2 space o n space C H subscript 3 N H subscript 2$ gives:

Action of $H N O subscript 2 space o n space C H subscript 3 N H subscript 2$ gives:

chemistry-General

Primary alcohols can be obtained from the reaction of the $R M g X$ with:

Primary alcohols can be obtained from the reaction of the $R M g X$ with:

chemistry-General

$R long dash C H equals C H subscript 2$ reacts with in presence of $H subscript 2 O subscript 2$ to give:

$R long dash C H equals C H subscript 2$ reacts with in presence of $H subscript 2 O subscript 2$ to give:

chemistry-General

parallel

The quadratic equation whose roots are I and m where l= $lim for theta not stretchy rightwards arrow 0 of open parentheses fraction numerator 3 sin space theta minus 4 sin cubed space theta over denominator theta end fraction close parentheses m equals lim for theta not stretchy rightwards arrow 0 of open parentheses fraction numerator 2 tan space theta over denominator theta open parentheses 1 minus tan squared space theta close parentheses end fraction close parentheses$ is

The quadratic equation whose roots are I and m where l= $lim for theta not stretchy rightwards arrow 0 of open parentheses fraction numerator 3 sin space theta minus 4 sin cubed space theta over denominator theta end fraction close parentheses m equals lim for theta not stretchy rightwards arrow 0 of open parentheses fraction numerator 2 tan space theta over denominator theta open parentheses 1 minus tan squared space theta close parentheses end fraction close parentheses$ is

$L subscript x not stretchy rightwards arrow 0 end subscript equals fraction numerator tan cubed begin display style space end style x minus sin cubed space x over denominator x to the power of 5 end fraction$

$L subscript x not stretchy rightwards arrow 0 end subscript equals fraction numerator tan cubed begin display style space end style x minus sin cubed space x over denominator x to the power of 5 end fraction$

$L subscript x not stretchy rightwards arrow 0 end subscript equals fraction numerator sin begin display style space end style open parentheses pi cos squared space x close parentheses over denominator x squared end fraction$ Is equal to

$L subscript x not stretchy rightwards arrow 0 end subscript equals fraction numerator sin begin display style space end style open parentheses pi cos squared space x close parentheses over denominator x squared end fraction$ Is equal to

parallel

$L t subscript x plus 0 end subscript fraction numerator left parenthesis 1 minus cos space 2 x right parenthesis left parenthesis 3 plus cos space x right parenthesis over denominator x tan space 4 x end fraction$ is equal to

$L t subscript x plus 0 end subscript fraction numerator left parenthesis 1 minus cos space 2 x right parenthesis left parenthesis 3 plus cos space x right parenthesis over denominator x tan space 4 x end fraction$ is equal to

$Lt subscript x not stretchy rightwards arrow 0 end subscript space fraction numerator tan begin display style space end style x minus sin space x over denominator x squared end fraction equals$

$Lt subscript x not stretchy rightwards arrow 0 end subscript space fraction numerator tan begin display style space end style x minus sin space x over denominator x squared end fraction equals$

$Lt subscript x not stretchy rightwards arrow 8 end subscript space fraction numerator square root of 1 plus square root of 1 plus x end root end root minus 2 over denominator x minus 8 end fraction equals$

$Lt subscript x not stretchy rightwards arrow 8 end subscript space fraction numerator square root of 1 plus square root of 1 plus x end root end root minus 2 over denominator x minus 8 end fraction equals$

parallel

$L t subscript x not stretchy rightwards arrow 0 end subscript fraction numerator x over denominator square root of x plus 4 end root minus 2 end fraction equals$

$L t subscript x not stretchy rightwards arrow 0 end subscript fraction numerator x over denominator square root of x plus 4 end root minus 2 end fraction equals$

$Lt subscript x not stretchy rightwards arrow 0 end subscript fraction numerator cube root of 1 plus x end root minus cube root of 1 minus x end root over denominator x end fraction equals$

$Lt subscript x not stretchy rightwards arrow 0 end subscript fraction numerator cube root of 1 plus x end root minus cube root of 1 minus x end root over denominator x end fraction equals$

If $f left parenthesis x right parenthesis equals negative square root of 25 minus x squared end root comma text then end text L t subscript x not stretchy rightwards arrow 1 end subscript fraction numerator f left parenthesis x right parenthesis minus f left parenthesis 1 right parenthesis over denominator x minus 1 end fraction equals$

If $f left parenthesis x right parenthesis equals negative square root of 25 minus x squared end root comma text then end text L t subscript x not stretchy rightwards arrow 1 end subscript fraction numerator f left parenthesis x right parenthesis minus f left parenthesis 1 right parenthesis over denominator x minus 1 end fraction equals$

parallel

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