#combinatorics

Question

How many ways the letters of the word “AABCCD” can be arranged such that, these neither begin with ‘A’ nor end with D ?

Comments

i m getting 168

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if total arrangements are $\frac{6!}{2!*2!}=180$ , then after restrictions how it can be 252?

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@Verma Ashish

i also have same doubt but total no. of words are given as 6!/2! = 360.

answer given is 360 - (60+60 -12) = 252.

 

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if you assume word as "AXBCCD" then $total-(starting\; with\; A\; or\; ending\; with\;D) $

then it results $360-108=252$.

X can be anything other than A,B,C,D.

Answer 1

Answer should be 102

Total arrangements are 180

Starting with A: 5!/2! = 60

Ending with D: 5!/2!*2! = 30

Starting with A & Ending with B = 4!/2! = 12

After the given restrictions it will be: 180-60-30+12 = 102

Comments

Then i get 90

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in case of A _ _ _ _ non D, you will get only 48 arrangements

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Yes you are right..👍