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Source: — Problem Solving |

by Bill@VeritasPrep » Mon May 07, 2012 8:43 am
a) If we set r in the second spot, there are 5 spots for 5 letters. 5!=120

b) We can put qe next to each other in the following ways:

q e _ _ _ _
_ q e _ _ _
_ _ q e _ _
_ _ _ q e _
_ _ _ _ q e

For each of these, we have 4 spots for 4 letters: 4!=24. 24*5=120. However, for each of the arrangements we identified, we can reverse the order of q and e, so we need to double that total. 120*2=240.
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by aneesh.kg » Mon May 07, 2012 10:02 am
We have these alphabets to use:
S,Q,U,A,R,E.

a) R has the be the second letter. The remaining 5 alphabets can be arranged on the remaining places in 5! = 120 ways.

b) Since Q and E have to together, lets first group them like this: (Q, E) and treat it as one single entity.
We have these 5 entities now to be arranged: (Q,E),S,U,A,R
They can be arranged in 5! ways. But, the Q and E can be arranged inside the bracket in 2! ways because the question has not mentioned that Q has to come before E.
Total number of ways = 5!*2! = 120*2 = 240
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