A person throws four standard six sided distinguishable dice. Number of ways in which he can throw if the product of the four number shown on the upper faces is 144, is

Question

A person throws four standard six sided distinguishable dice. Number of ways in which he can throw if the product of the four number shown on the upper faces is 144, is (A) 24 (B) 36 (C) 42 (D) 48

Tardigrade Admin · Accepted Answer

Possible cases if the product of four numbers a ⋅ b ⋅ c ⋅ d=144(1 ≤ a, b, c, d ≤ 6) 6,6,2,2, ; 6,6,4,1 ; 6,4,3,2 and 4,4,3,3 =(4 !/2 ! ⋅ 2 !)+(4 !/2 !)+4 !+(4 !/2 ! ⋅ 2 !)=48

Q. A person throws four standard six sided distinguishable dice. Number of ways in which he can throw if the product of the four number shown on the upper faces is 144, is

24

36

42

48

Possible cases if the product of four numbers a $\cdot b \cdot c \cdot d = 144 (1 \leq a, b, c, d \leq 6)$
$6, 6, 2, 2,; 6, 6, 4, 1; 6, 4, 3, 2$ and $4, 4, 3, 3$
$= \frac{4 !}{2 ! \cdot 2 !} + \frac{4 !}{2 !} + 4! + \frac{4 !}{2 ! \cdot 2 !} = 48$

Q. A person throws four standard six sided distinguishable dice. Number of ways in which he can throw if the product of the four number shown on the upper faces is 144, is

24

36

42

48

Possible cases if the product of four numbers a ⋅b⋅c⋅d=144(1≤a,b,c,d≤6) 6,6,2,2,;6,6,4,1;6,4,3,2 and 4,4,3,3 =2!⋅2!4!​+2!4!​+4!+2!⋅2!4!​=48

Possible cases if the product of four numbers a $\cdot b \cdot c \cdot d = 144 (1 \leq a, b, c, d \leq 6)$
$6, 6, 2, 2,; 6, 6, 4, 1; 6, 4, 3, 2$ and $4, 4, 3, 3$
$= \frac{4 !}{2 ! \cdot 2 !} + \frac{4 !}{2 !} + 4! + \frac{4 !}{2 ! \cdot 2 !} = 48$