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The number of ways in which three persons can dress themselves when they have 4 shirts. 5 pants and 6 hats between them, is-

⁴C₃ × ⁵C₃ ×⁶C₃
⁴P₃×⁵P₃ × ⁶P₃
$\frac{15!}{4! 5! 6!}$
$\frac{15!}{(3!)^{3}}$

hint Hint:

Here, we can use the formula
$C presuperscript n subscript r space cross times r factorial equals space fraction numerator n factorial over denominator left parenthesis n minus r factorial right parenthesis end fraction space space equals P presuperscript n subscript r$

The correct answer is: ⁴P₃×⁵P₃ × ⁶P₃

Given that,
Shirts = 4 , Pants = 5 and Hats = 6, which are distributed among 3 men.
For shirts, Number of ways they can wear = $C presuperscript 4 subscript 3 space end subscript cross times 3 factorial$
For pants, Number of ways they can wear = $C presuperscript 5 subscript 3 cross times 3 factorial$
For hats, number of ways they can wear = $C presuperscript 6 subscript 3 cross times 3 factorial$
Total number of ways they can wear
$rightwards double arrow left parenthesis C presuperscript 4 subscript 3 cross times 3 factorial right parenthesis cross times left parenthesis C presuperscript 5 subscript 3 cross times 3 factorial right parenthesis cross times left parenthesis C presuperscript 6 subscript 3 cross times 3 factorial right parenthesis rightwards double arrow fraction numerator 4 factorial over denominator 3 factorial thin space 1 factorial end fraction cross times 3 factorial cross times thin space fraction numerator 5 factorial over denominator 3 factorial thin space 2 factorial end fraction cross times 3 factorial cross times thin space fraction numerator 6 factorial over denominator 3 factorial thin space 3 factorial end fraction cross times 3 factorial rightwards double arrow fraction numerator 4 factorial over denominator 1 factorial end fraction cross times fraction numerator 5 factorial over denominator 2 factorial end fraction cross times fraction numerator 6 factorial over denominator 3 factorial end fraction space left parenthesis C presuperscript n subscript r space equals space fraction numerator n factorial over denominator left parenthesis n minus r factorial right parenthesis end fraction space space equals P presuperscript n subscript r right parenthesis rightwards double arrow P presuperscript 4 subscript 3 cross times P presuperscript 5 subscript 3 cross times P presuperscript 6 subscript 3$

Thus, the number of ways in which three persons can dress themselves when they have 4 shirts. 5 pants and 6 hats between them, is $P presuperscript 4 subscript 3 cross times P presuperscript 5 subscript 3 cross times P presuperscript 6 subscript 3$ .

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The number of ways in which three persons can dress themselves when they have 4 shirts. 5 pants and 6 hats between them, is-

4C3 × 5C3 ×6C3 4P3× 5 P3 × 6 P3

Here, we can use the formula

The correct answer is: 4P3× 5 P3 × 6 P3

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⁴C₃ × ⁵C₃ ×⁶C₃
⁴P₃×⁵P₃ × ⁶P₃
$\frac{15!}{4! 5! 6!}$
$\frac{15!}{(3!)^{3}}$

Here, we can use the formula
$C presuperscript n subscript r space cross times r factorial equals space fraction numerator n factorial over denominator left parenthesis n minus r factorial right parenthesis end fraction space space equals P presuperscript n subscript r$

The correct answer is: ⁴P₃×⁵P₃ × ⁶P₃

If ^(m+n)P₂ = 56 and ^m–nP₂ = 12 then (m, n) equals-

If ^(m+n)P₂ = 56 and ^m–nP₂ = 12 then (m, n) equals-

A thin uniform annular disc (see figure) of mass $M$ has outer radius $4 R$ and inner radius $3 R$ . The work required to take a unit mass from point $P$ on its axis to infinity is

A thin uniform annular disc (see figure) of mass $M$ has outer radius $4 R$ and inner radius $3 R$ . The work required to take a unit mass from point $P$ on its axis to infinity is

The coefficient of $x to the power of negative n end exponent$ in $left parenthesis 1 plus x right parenthesis to the power of n end exponent open parentheses 1 plus fraction numerator 1 over denominator x end fraction close parentheses to the power of n end exponent$ is

The coefficient of $x to the power of negative n end exponent$ in $left parenthesis 1 plus x right parenthesis to the power of n end exponent open parentheses 1 plus fraction numerator 1 over denominator x end fraction close parentheses to the power of n end exponent$ is

If one root of the equation $a x squared plus b x plus c equals 0$ is reciprocal of the one of the roots of equation $a subscript 1 x squared plus b subscript 1 x plus c subscript 1 equals 0$ then

If one root of the equation $a x squared plus b x plus c equals 0$ is reciprocal of the one of the roots of equation $a subscript 1 x squared plus b subscript 1 x plus c subscript 1 equals 0$ then

If the quadratic equation $a x squared plus 2 c x plus b equals 0$ and $a x squared plus 2 b x plus c equals 0 left parenthesis b not equal to c right parenthesis$ have a common root then $a plus 4 b plus 4 c$ is equal to

If the quadratic equation $a x squared plus 2 c x plus b equals 0$ and $a x squared plus 2 b x plus c equals 0 left parenthesis b not equal to c right parenthesis$ have a common root then $a plus 4 b plus 4 c$ is equal to