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A shopkeeper places before you 41 different toys out of which 20 toys are to be purchased. Suppose m= number of ways in which 20 toys can be purchased without any restriction and n= number of ways in which a particular toy is to be always included in each selection of 20 toys, then (m-n) can be expressed as:

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$(\frac{21}{1}) (\frac{22}{2}) (\frac{23}{3}) ..... (\frac{40}{20})$

$\frac{2^{20} (1.3.5.....19)}{10!}$

$\prod_{r = 0}^{19} (\frac{4 r + 2}{20 - r})$

$\frac{2^{10}}{20!} (1.3.5....39)$

answer is C, D.

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Detailed Solution

$m =^{41} C_{20}; n =^{40} C_{19}$

$m - n =^{41} C_{20} -^{40} C_{19}$

$=^{40} C_{20} +^{40} C_{19} -^{40} C_{19}$

$=^{40} C_{20} = \frac{40!}{20! .20!}$

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