The number of ways of seven digit number with distinct digits of the form `a_(1)a_(2)a_(3)a_(4)a_(5)a_(6)a_(7),(a_9(i)!=0AAi=1,2,.7)` be present in decimal system such that `a_(1)lta_(2)lta_(3)lta_(4)gta_(5)gta_(6)gta_(7)`

Options

A. `820`
B. `720`
C. `620`
D. `120`

Explanation

To solve the problem of finding the number of ways to form a seven-digit number with distinct digits such that the digits satisfy the conditions \( a_1 < a_2 < a_3 < a_4 > a_5 > a_6 > a_7 \), we can follow these steps:

### Step 1: Identify the largest digit
Since \( a_4 \) is the peak of the sequence (the largest digit), we can choose \( a_4 \) to be one of the digits from 7 to 9. This is because \( a_1, a_2, a_3 \) must be less than \( a_4 \) and \( a_5, a_6, a_7 \) must also be less than \( a_4 \).

### Step 2: Case analysis based on the value of \( a_4 \)
We will consider three cases based on the value of \( a_4 \):

1. **Case 1: \( a_4 = 7 \)**
- The digits less than 7 are {0, 1, 2, 3, 4, 5, 6} (total 6 digits).
- We need to choose 3 digits from these 6 for \( a_1, a_2, a_3 \) (which must be in increasing order).
- The remaining 3 digits will be \( a_5, a_6, a_7 \) (which must be in decreasing order).
- The number of ways to choose 3 digits from 6 is \( \binom{6}{3} \).
- The number of ways to arrange the remaining 3 digits in decreasing order is \( \binom{3}{3} = 1 \).

Total ways for Case 1:
\[ \binom{6}{3} \times 1 = 20 \]

2. **Case 2: \( a_4 = 8 \)**
- The digits less than 8 are {0, 1, 2, 3, 4, 5, 6, 7} (total 7 digits).
- Choose 3 digits from these 7 for \( a_1, a_2, a_3 \).
- The remaining 4 digits will be \( a_5, a_6, a_7 \).
- The number of ways to choose 3 digits from 7 is \( \binom{7}{3} \).
- The number of ways to arrange the remaining 4 digits in decreasing order is \( \binom{4}{3} = 4 \).

Total ways for Case 2:
\[ \binom{7}{3} \times \binom{4}{3} = 35 \times 4 = 140 \]

3. **Case 3: \( a_4 = 9 \)**
- The digits less than 9 are {0, 1, 2, 3, 4, 5, 6, 7, 8} (total 8 digits).
- Choose 3 digits from these 8 for \( a_1, a_2, a_3 \).
- The remaining 5 digits will be \( a_5, a_6, a_7 \).
- The number of ways to choose 3 digits from 8 is \( \binom{8}{3} \).
- The number of ways to arrange the remaining 5 digits in decreasing order is \( \binom{5}{3} = 10 \).

Total ways for Case 3:
\[ \binom{8}{3} \times \binom{5}{3} = 56 \times 10 = 560 \]

### Step 3: Sum the total ways from all cases
Now, we sum the total ways from all three cases:
\[ 20 + 140 + 560 = 720 \]

Thus, the total number of ways to form the seven-digit number is **720**.