Maths-

General

Easy

Question

If $alpha$ = mC₂, then $alpha$ C₂ is equal to -

${}^{m + 1} C_{4}$
${}^{m - 1} C_{4}$

${}^{3 m + 2} C_{4}$
${}^{3 m + 1} C_{4}$

hint Hint:

We will expand the value of $α =^{m} C_{2}$ using $^{n} C_{r} = \frac{n!}{r! (n - r)!}$ . Similarly, simplify $^{α} C_{2}$ and then substitute the value of $α$ in the expression of $^{α} C_{2}$ . Then, use the property of factorial and simplify the expression to get the required answer.

The correct answer is: $space presuperscript 3 m plus 1 end presuperscript C subscript 4$

$W e space a r e space g i v e n space t h a t comma space alpha space equals space scriptbase C subscript 2 end scriptbase presuperscript m$
$W e space w i l l space e x p a n d space t h e space e x p r e s s i o n space b y space u sin g space t h e space i d e n t i t y space scriptbase C subscript r space equals space fraction numerator n factorial over denominator r factorial space left parenthesis n minus r right parenthesis factorial end fraction end scriptbase presuperscript n$

$H e n c e comma space alpha equals fraction numerator m factorial over denominator 2 factorial left parenthesis m minus 2 right parenthesis factorial end fraction$
$I t space i s space a l s o space k n o w n space t h a t space space n factorial equals n. left parenthesis n minus 1 right parenthesis. left parenthesis n minus 2 right parenthesis.....3.2.1$

$T h e n comma space alpha equals fraction numerator m left parenthesis m minus 1 right parenthesis left parenthesis m minus 2 right parenthesis factorial over denominator 2.1 left parenthesis m minus 2 right parenthesis factorial end fraction rightwards double arrow alpha equals fraction numerator m left parenthesis m minus 1 right parenthesis over denominator 2 end fraction$

$W e space h a v e space t o space f i n d space t h e space v a l u e space o f space scriptbase C subscript 2 space end scriptbase presuperscript alpha w h i c h space i s space a l s o space e q u a l space t o space fraction numerator space alpha factorial over denominator 2 factorial left parenthesis alpha minus 2 right parenthesis factorial end fraction$
$A l s o comma space w e space c a n space r e w r i t e space i t space a s space fraction numerator alpha left parenthesis alpha minus 1 right parenthesis left parenthesis alpha minus 2 right parenthesis factorial over denominator 2 factorial left parenthesis alpha minus 2 right parenthesis end fraction space equals space fraction numerator alpha left parenthesis alpha minus 1 right parenthesis over denominator 2 end fraction O n space s u b s t i t u t i n g space t h e space v a l u e s space o f space alpha comma space w e space w i l l space g e t comma fraction numerator m. left parenthesis m minus 1 right parenthesis over denominator 2 end fraction left parenthesis fraction numerator m. left parenthesis m minus 1 right parenthesis over denominator 2 end fraction minus 1 right parenthesis space equals space fraction numerator m. left parenthesis m minus 1 right parenthesis left parenthesis m. left parenthesis m minus 1 right parenthesis minus 2 right parenthesis over denominator 8 end fraction S o l v e space t h e space b r a c k e t s. fraction numerator left parenthesis m squared minus m right parenthesis left parenthesis m squared minus m minus 2 right parenthesis over denominator 8 end fraction equals space fraction numerator left parenthesis m squared minus m right parenthesis left parenthesis m squared minus 2 m space plus space m minus 2 right parenthesis over denominator 8 end fraction equals space fraction numerator left parenthesis m left parenthesis m minus 1 right parenthesis left parenthesis m space left parenthesis m minus 2 right parenthesis space plus 1 left parenthesis space m minus 2 right parenthesis right parenthesis over denominator 8 end fraction equals space fraction numerator left parenthesis m minus 2 right parenthesis left parenthesis m minus 1 right parenthesis m left parenthesis m plus 1 right parenthesis over denominator 8 end fraction W e space w i l l space m u l t i p l y space a n d space d i v i d e space b y space left parenthesis m minus 3 right parenthesis factorial fraction numerator space left parenthesis m minus 2 right parenthesis left parenthesis m minus 1 right parenthesis m left parenthesis m plus 1 right parenthesis left parenthesis m minus 3 right parenthesis factorial over denominator 8 left parenthesis m minus 3 right parenthesis factorial end fraction space equals space fraction numerator left parenthesis m plus 1 right parenthesis factorial over denominator 8 left parenthesis m minus 3 right parenthesis factorial end fraction W e space c a n space w r i t e space m minus 3 space a s space left parenthesis left parenthesis m plus 1 right parenthesis minus 4 right parenthesis T h e n comma fraction numerator left parenthesis m plus 1 right parenthesis factorial space space over denominator 8 left parenthesis left parenthesis left parenthesis m plus 1 right parenthesis minus 4 right parenthesis right parenthesis factorial end fraction N e x t comma space m u l t i p l y space a n d space d i v i d e space b y space 3 space t o space m a k e space e x p r e s s i o n space 4 factorial space I n space t h e space d e n o m i n a t o r. fraction numerator 3 left parenthesis m plus 1 right parenthesis factorial over denominator 4 factorial left parenthesis left parenthesis left parenthesis m plus 1 right parenthesis minus 4 right parenthesis right parenthesis factorial end fraction equals scriptbase C subscript 4 end scriptbase presuperscript 3 m plus 1 end presuperscript$

Students must know how to open expressions like $^{n} C_{r}$ .
Also, the value of $n!$ is $n . (n - 1) . (n - 2) . . . . .3 .2 .1$ .
Combination is used to select $r$ objects from $n$ objects when the arrangement of the objects does not matter.

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If = mC2, then C2 is equal to -

We will expand the value of α=mC2 using nCr=n!r!(n−r)!. Similarly, simplify αC2 and then substitute the value of α in the expression of αC2. Then, use the property of factorial and simplify the expression to get the required answer.

The correct answer is:

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If $alpha$ = mC₂, then $alpha$ C₂ is equal to -

We will expand the value of $α =^{m} C_{2}$ using $^{n} C_{r} = \frac{n!}{r! (n - r)!}$ . Similarly, simplify $^{α} C_{2}$ and then substitute the value of $α$ in the expression of $^{α} C_{2}$ . Then, use the property of factorial and simplify the expression to get the required answer.

The correct answer is: $space presuperscript 3 m plus 1 end presuperscript C subscript 4$

Given the circuit :

Which of the following is equivalent to the above circuit ?
I)
II)
The true statement in the above is / are

Given the circuit :

Which of the following is equivalent to the above circuit ?
I)
II)
The true statement in the above is / are

Enthalpy change for the reaction, 4H_(g) →2H_2(g) is–869.6kJ The dissociation energy of H–H bond is

Enthalpy change for the reaction, 4H_(g) →2H_2(g) is–869.6kJ The dissociation energy of H–H bond is

Assertion (A) : The number of ways of arranging 6 boys and 5 girls alternately at circular table is $0$ Reason (R) : To arrange boys and girls alternately at a circular table, they should be equal in number.

Assertion (A) : The number of ways of arranging 6 boys and 5 girls alternately at circular table is $0$ Reason (R) : To arrange boys and girls alternately at a circular table, they should be equal in number.