To solve the problem, we need to determine the number of seven-digit numbers in set \( P \) where each digit is from the set \( \{1, 2, 3\} \), and the sum of the digits equals 11. Let \( d_1 d_2 d_3 d_4 d_5 d_6 d_7 \) represent the seven-digit number, with the following conditions:
\( d_1 + d_2 + d_3 + d_4 + d_5 + d_6 + d_7 = 11 \)
Let \( n_1 \), \( n_2 \), and \( n_3 \) represent the number of times the digits 1, 2, and 3 appear, respectively. The following constraints apply:
1. Simplifying the Equations:
From the first equation, \( n_1 = 7 - n_2 - n_3 \). Substituting this into the second equation:
\( (7 - n_2 - n_3) + 2n_2 + 3n_3 = 11 \)
\( 7 + n_2 + 2n_3 = 11 \)
\( n_2 + 2n_3 = 4 \)
2. Finding Non-Negative Integer Solutions:
We now solve \( n_2 + 2n_3 = 4 \) for non-negative integers \( n_2 \) and \( n_3 \):
If \( n_3 > 2 \), then \( 2n_3 > 4 \), which would make \( n_2 \) negative. Thus, there are no other valid solutions.
3. Total Number of Elements in Set \( P \):
The total number of elements in set \( P \) is the sum of the number of arrangements for all cases:
\( 35 + 105 + 21 = 161 \).
Final Answer:
The final answer is \( \boxed{161} \).
If \[ \sum_{r=1}^{30} r^2 \left( \binom{30}{r} \right)^2 = \alpha \times 2^{29}, \] then \( \alpha \) is equal to _______.
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,