Question

53. Find the number of ways in which arrangements of 4 letters can be made from the word "MATHEMATICS. (a) 1,680 (b) 756 (C) 18 (d) 2454

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Maths

Solution

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In marthematics ( n O cdot ) of ( r A=2 ) no. of ( T=2 ) no. ( A=2 ) ofter luters ( forall, E, I, c, s ) are sinfle letters alike ond two alik case ( 1-T w_{0} ) We have 3 groups of same luters ( i cdot e cdot 2 M cdot s, 2 T^{prime} s ) ard ( 2 A_{S}^{prime} ) ad we have to sulet two out so this can be dom in ( 3 C_{2} ) two different. here ir this owt of 7 group ( i cdot f cdotleft(2 text { pes } 27^{circ} ) sent. right. ( 7 c_{2}+3 c_{1} ) ( mid cos x-3] ) all diffirent differet letors ( quad M, T, A_{1} H, E, I, C S ) this cun be don in 8 cumay Arrarginent - for ist con. ( frac{4 !}{2 ! 2 !} ) for ( i^{-d} cos frac{4 !}{2 !} ) dor third con cil. final anw ( -3 c_{2} frac{x q !}{2 ! 2 !}+3 c_{1} times 7 c_{2} times frac{4 !}{2 !}+8 c_{4} times 4 ! ) ( 18+756+1680=2454 )