A baby presses the ten numbers (zero through nine) on a phone dial pad once each. How many different number sequences could she have dialed?

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A baby presses the ten numbers (zero through nine) on a phone dial pad once each. How many different number sequences could she have dialed?

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Answer 1 · 2017-03-17T02:29:30

$10! = 3, 628, 800$ $10! = 3,628,800$

Explanation:

The first number the baby presses could be any of the 10 keys - which gives us 10 choices.

The second number pushed can be any of the 9 remaining keys - and so 9 choices.

The third number pushed can be any of the 8 remaining keys - and so 8 choices.

And so on, down to the last key. We can multiply all these choices together:

$10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 10! = 3, 628, 800$ $10xx9xx8xx7xx6xx5xx4xx3xx2xx1=10! = 3,628,800$

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A baby presses the ten numbers (zero through nine) on a phone dial pad once each. How many different number sequences could she have dialed?

1 Answer

Explanation:

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