Combinatorics

Question

Three  ladies  have  each  brought  a  child  for  admission  to  a  school. The  head  of  the  school  wishes  to  interview  the  six  people  one  by  one, taking  care  that  no  child  is  interviewed  before  its  mother.  In  how many different  ways  can  the  interviews be  arranged?

• A. $6$
• B. $36$
• C. $72$
• D. $90$

Answer 1

This is very much similar to Balanced parenthesis consider ( as mother and ) as child

total possible permutations

1. $MCMCMC$ =   $3!$ = $6$

2. $MCMMCC$ = $\binom{3}{1} * 4$ = $12$ (selecting first pair MC out of 3 pairs and then for each 4 combinations of MMCC)

3. $MMCCMC$ = $ 4*\binom{3}{1}$ = $12$ (same as above)

4. $MMMCCC$ = $3!$ *$3!$ = $36$

5. $MMCMCC$ = $3 *2*2*1*2*1$ = $24$

6+12+12+36+24 = $90$

Comments

I havenot got point 2

how 4 combination of MMCC be 4?

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@srestha di, in MMCC, the M's are distinct so we can alter their positions in 2! ways and also the children in 2! ways, hence= 2!*2!=4

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$M_{1}M_{2}C_{1}C_{2}$

$M_{1}M_{2}C_{2}C_{1}$

$M_{2}M_{1}C_{1}C_{2}$

$M_{2}M_{1}C_{2}C_{1}$

and selecting first MC, 1 out of 3 pairs of mom and child

Answer 2

Let us denote the mothers by M and children by C. M1 is mother of C1, M2->C2, M3->C3. Now, we need to ensure that M1, C1 are in order, M2, C2 in order and M3 , C3 are in order. Out of 6 places, choose 2 for M1,C1=$\binom{6}{2}$, then out of remaining 4 places choose 2 for M2,C2 then $\binom{4}{2}$ and rest 2 places can be occupied in 1 way by M3,C3.

$\therefore$ Total no of ways=$\binom{6}{2}*\binom{4}{2}=15*6=90$

Answer 3

90 ways as 6!/2!2!2!

Answer 4

It should be 6C2*4C2*2C2= 90

Comments

please explain

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please explain how you get the answer