To solve the problem, we need to count the number of 5-digit numbers $d_1d_2d_3d_4d_5$ where each digit $d_i$ is from the set {0, 1, 2, 3, 4, 5, 6, 7}, the number is greater than 50000, and $d_1 + d_5 \leq 8$.
1. Determine the possible values for $d_1$:
Since the number must be greater than 50000, $d_1$ can only be 5, 6, or 7. Thus, $d_1 \in \{5, 6, 7\}$.
2. Analyze the constraint $d_1 + d_5 \leq 8$ for each possible value of $d_1$:
3. Determine the number of possibilities for $d_2$, $d_3$, and $d_4$:
Since there are no restrictions on $d_2$, $d_3$, and $d_4$ other than belonging to the set {0, 1, 2, 3, 4, 5, 6, 7}, each of them can take 8 possible values. Therefore, there are $8 \times 8 \times 8 = 8^3 = 512$ possibilities for $d_2d_3d_4$.
4. Calculate the total number of such 5-digit numbers:
We consider each case for $d_1$ separately and sum the results:
Therefore, the total number of such 5-digit numbers is $2048 + 1536 + 1024 = 4608$.
Final Answer:
The total number of such 5 digit numbers is $ {4608} $.
Match List-I with List-II
List-I | List-II |
---|---|
(A) \(^{8}P_{3} - ^{10}C_{3}\) | (I) 6 |
(B) \(^{8}P_{5}\) | (II) 21 |
(C) \(^{n}P_{4} = 360,\) then find \(n\). | (III) 216 |
(D) \(^{n}C_{2} = 210,\) find \(n\). | (IV) 6720 |
Choose the correct answer from the options given below:
If $ \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p $, then $ 96 \log_e p $ is equal to _______