A company bought for its 7 offices 2 computers of brand N and 3 computers of brand M. In how many ways could computers be distributed among the offices if each office can have only 1 computer. Options:
A) 196  B) 210  C) 256  D) 292  E) 312
I solved it in a slightly different way. I would love feedback on the logic.   Again, this is a comb problem. So I though of it as follows.   From the seven offices, there will be 5 'winners' (of PCs) and 2'losers'. The number of ways they could be winners and losers is 7!/(5!2!) = 21   For the five winners, the 5 computers can be distributed in 5!/(2!3!) ways = 10.   If you have trouble seeing this, think of anagrams NNMMM (so 5! ways of ordering them but divide by 2! because 2 Ns are similar and 3! because 3 'M's are similar.)   The number of different combinations is then 21*10= 
<gadget id="calculator">21*10</gadget>
<output>210</output>
210 -&gt; 210.
<result>B</result>
A company bought for its 7 offices 2 computers of brand N and 3 computers of brand M. In how many ways could computers be distributed among the offices if each office can have only 1 computer.