PnC 1. (m+n)P2=56 & (m-n)P2 =12 FIND M AND N ... (WITH EXPLANATION ) 2. IN HOW MANY WAYS 6 EXAMINATION Q. PAPER OUT OF WHICH 2 ARE MATHS CAN BE ARRANGEND SO THAT 2 MATHEMATICS PAPER NEVER COM TOGETHER ? (WITH EXPLANATION ) 3.IN HOW MANY WAYS 3 BOYS AND 5 GIRLS BE ARRANGED IN A ROW SO THAT i)NO TWO BOYS R TOGETHER ii)ALL THE GILS ARE TOGETHER.(WITH EXPLANATION ) 4.FIND THE PERMUTATION OF THE LETTER INSTITUTION i) IN HOW MANY WAYS 3 I's COME TOGETHER (WITH EXPLANATION ) 5. IN HOW MANY WAYS 6 BOYS AND 6 GIRLS BE ARRANGED FOR A DINNER TABLE IF NO SIT TOGETHER (WITH EXPLANATION )

Asked by Satyam sinha | 13th Feb, 2014, 12:08: PM

Expert Answer:

You are requested to ask one question at a time. The solutions to your first two questions are as follows:
 
Solution 1:
P(m+n, 2) = (m+n)(m+n-1) = 56 = 8 * 7,
Thus, m+n = 8 ....(1)
and P(m-n, 2) = (m-n)(m-n-1) = 12 = 4 * 3,
Thus, m - n = 6 .....(2)

Adding (1) and (2), we get
2m = 8+4 = 12
i.e. m = 6

From (1), n = 8 - m = 8 - 6 = 2
 
Solution 2:
 6  papers can be arranged in 6! ways.
When 2 mathematics paper are together, the number of arrangements will be 5! x 2.
When 2 mathematics papers are not together, the number of arrangements will be
6! - (5!)2 = 5!(6 - 2) = (4) x(5!) = 480 ways

Answered by  | 13th Feb, 2014, 02:25: PM

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