Four faces of a tetrahedral dice are marked with 2, 3, 4, 5. The lowest face is considered as a outcome.The number of ways in which a total of 33 can occur in 7 throws, is equal to

Question

Four faces of a tetrahedral dice are marked with 2, 3, 4, 5. The lowest face is considered as a outcome.The number of ways in which a total of 33 can occur in 7 throws, is equal to (A) 28 (B) 7 (C) 35 (D) 42

Tardigrade Admin · Accepted Answer

5555553 ⇒ (7 !/6 !)=7 5555544 ⇒ (7 !/5 ! 2 !)=21 ∴ text Total ways =28

Q. Four faces of a tetrahedral dice are marked with 2, 3, 4, 5. The lowest face is considered as a outcome.The number of ways in which a total of 33 can occur in 7 throws, is equal to

28

7

35

42

$5555553 \Rightarrow \frac{7 !}{6 !} = 7$
$5555544 \Rightarrow \frac{7 !}{5 ! 2 !} = 21$
$∴ Total ways = 28$

Q. Four faces of a tetrahedral dice are marked with 2, 3, 4, 5. The lowest face is considered as a outcome.The number of ways in which a total of 33 can occur in 7 throws, is equal to

28

7

35

42

5555553⇒6!7!​=7 5555544⇒5!2!7!​=21 ∴ Total ways =28

$5555553 \Rightarrow \frac{7 !}{6 !} = 7$
$5555544 \Rightarrow \frac{7 !}{5 ! 2 !} = 21$
$∴ Total ways = 28$