The number of ways in which the following pri | vedclass

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(A) There are $6$ prizes in total: $1^{st}$ Math,$2^{nd}$ Math,$1^{st}$ Physics,$2^{nd}$ Physics,$1^{st}$ Chemistry,and $1^{st}$ English.First,conside

Vedclass · Accepted Answer

$20^4 	imes 19^2$

Vedclass · Answer

(A) There are $6$ prizes in total: $1^{st}$ Math,$2^{nd}$ Math,$1^{st}$ Physics,$2^{nd}$ Physics,$1^{st}$ Chemistry,and $1^{st}$ English.First,consider the $4$ first prizes ($1^{st}$ Math,$1^{st}$ Physics,$1^{st}$ Chemistry,$1^{st}$ English). Each of these can be awarded to any of the $20$ boys. Since one boy can win multiple first prizes,the number of ways to award these $4$ prizes is $20 	imes 20 	imes 20 	imes 20 = 20^4$.Next,consider the $2$ second prizes ($2^{nd}$ Math,$2^{nd}$ Physics). $A$ boy who has already won a first prize in a subject cannot win the second prize in that same subject. Therefore,for the $2^{nd}$ Math prize,there are $20 - 1 = 19$ choices,and for the $2^{nd}$ Physics prize,there are $20 - 1 = 19$ choices.The total number of ways is $20^4 	imes 19 	imes 19 = 20^4 	imes 19^2$.

The number of ways in which the following prizes can be given to a class of $20$ boys,first and second Mathematics,first and second Physics,first Chemistry,and first English is:

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