A die is rolled 10 times.find th no. of ways so that outcomes always contain 1,2,3.
I cannot understand how to approach the solution.someone pls help.
THANKING YOU
Hello, ananya jana!
I'll take a guess at what the question asks . . .
Each roll has 6 possible outcomes.A die is rolled 10 times.
Find the number of ways so that outcomes always contain a 1, 2, or 3.
With a sequence of 10 rolls, there are: .$\displaystyle 6^{10}$ possible outcomes.
How many of these rolls contain no 1s, 2s, or 3s?
The rolls must consist of 4, 5, and/or 6 only.
. . There are: .$\displaystyle 3^{10}$ ways.
Therefore, there are: .$\displaystyle 6^{10} - 3^{19}\:=\:60,\!407,\!127$ ways.
but we cannot count the outcome where 1 4 5 6 4 5 4 5 6 1
because 2 and 3 are absent...
but this outcome may come in the
$\displaystyle 6^{10} - 3^{19}\:=\:60,\!407,\!127$
outcomes
Hello again, ananya jana!
See? . . . You have a different interpretation of the problem!
If that was the exact wording, the problem is poorly written.
A die is rolled 10 times.
Find the number of ways so that the outcomes
always contain at least one 1, one 2, and one 3.
There are: .$\displaystyle 6{10} = 60,\!466,\!176$ possible outcomes.
Let: .$\displaystyle \begin{array}{ccc}N1 &=& \text{no 1's} \\ N2 &=& \text{no 2's} \\ N3 &=& \text{no 3's} \end{array}$
. . $\displaystyle \begin{array}{c|c}
\text{Case} & \text{No. of ways} \\ \hline
N1 & 5^{10} \\ N2 & 5^{10} \\ N3 & 5^{10} \end{array}$
. . $\displaystyle \begin{array}{c|c}
\text{Case} & \text{No. of ways} \\ \hline
N1 \cap N2 & 4^{10} \\ N2 \cap N3 & 4^{10} \\ N1 \cap N3 & 4^{10} \end{array}$
. . $\displaystyle \begin{array}{c|c}
\text{Case} & \text{No. of ways} \\ \hline
N1 \cap N2 \cap N3 & 3^{10} \end{array}$
Formula: .$\displaystyle n(N1 \cup N2 \cup N3)$
. . $\displaystyle =\;n(N1) + n(N2) + n(N3) -$ $\displaystyle \bigg[n(N1\cap N2) + n(N2\cap N3) + n(N1\cap N3)\bigg] + n(N1 \cap N2 \cap N3) $
. . $\displaystyle =\;3\cdot5^{10} - 3\cdot4^{10} + 3^{10} \;=\;26,210,196 $
There are 26,210,196 outcomes with no 1 or no 2 or no 3.
Therefore, there are: .$\displaystyle 60,\!466,\!176 - 26,\!210,\!196 \:=\:\boxed{34,\!255,\!980}$ outcomes
. . that contain at least one 1, one 2, and one 3.
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Another interpretation . . .
Number of outcomes that contain exactly one 1, one 2, and one 3.
The 1,2,3 can occur in any of: .$\displaystyle 10\cdot9\cdot8 \:=\:720$ sequences of rolls.
The other seven rolls must not be 1, 2, or 3: .$\displaystyle 3^7$ ways.
Therefore, there are: .$\displaystyle 720 \times 3^7 \:=\:\boxed{1,\!574,\!640}$ ways.