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The total number of 8-digit numbers that contain only the 7-digits 2, 4, 5, 6, 7, 8, and 9 is

Option: 1

141120


Option: 2

132110


Option: 3

141110

 


Option: 4

131120


Answers (1)

best_answer

Given that,

There are 8-digit numbers that contain only the 7-digits 2, 4, 5, 6, 7, 8, and 9.

Choosing all the 7-digits is,

^{^{7}}C_{_{7}}^{^{7}}C_{_{7}}=1^{^{7}}C_{_{7}}=1ways

Now, in order to make the number 7-digits, we must choose one more digit from the 2, 4, 5, 6, 7, 8, and 9.

This is done in ^{^{7}}C_{_{1}}=7ways

The total number of permutations given by \frac{8 !}{2 !}

Thus, the required number of ways is,

\begin{aligned} & { }^7 C_7 \times{ }^7 C_1 \times \frac{8 !}{2 !}=1 \times 7 \times 20160 \\ & { }^7 C_7 \times{ }^7 C_1 \times \frac{8 !}{2 !}=141120 \end{aligned}

Therefore, the total number of ways is 141120.

 

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