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Re: A 6-digit number comprises of only 2’s and 3’s. How many of [#permalink]
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Last two digits should be 32. Rest 4 digits should be 2233 form ----> 4!/(2!*2!) = 6 = answer
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Re: A 6-digit number comprises of only 2’s and 3’s. How many of [#permalink]
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azl
Hello,

I have a silly question, how do you guys know that the 4 first number must include two 2's and two 3's?

Thank you in advance


Hi,

We have 32 A’s last two digits.
The first 4 can be any of 3 and 2, but all 6 when added should give you a multiple of 3, a property for multiple of 3.
Now you can have as many 3s as they are multiple of 3, but when you have a 2, either you have a 1 with it as 2+1=3 but that is not possible. So you have to have three or six etc of 2. Here you can have three 2s and one 2 is already used in units digit abcd32. So ABCD can have two 2s and the remaining will be 3s

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Re: A 6-digit number comprises of only 2s and 3s. How many of [#permalink]
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Re: A 6-digit number comprises of only 2s and 3s. How many of [#permalink]
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