Mathematics - Permutation & Combination Question with Solution | TestHub

MathematicsPermutation & CombinationnPr & nCrMedium2 minPYQ_2020

MathematicsMediumnumerical

The number of $4$ letter words (with or without meaning) that can be formed from the eleven letters of the word $EXAMINATION$ is

Answer:

2454.00

Solution:

$EXAMINATION$
$2 N, 2 A, 2 I, E, X, M, T, O$
Case $I :$ All are different, so $^{8} P_{4} = \frac{8!}{4!} = 8 \cdot 7 \cdot 6 \cdot 5 = 1680$
Case $I I :$ $2$ same and $2$ different, so $^{3} C_{1} {}^{7}C_{2} \frac{4!}{2!} = 3 \cdot 21 \cdot 12 = 756$
Case $I I I :$ $2$ same and $2$ same, so $^{3} C_{2} \frac{4!}{2! 2!} = 3 \cdot 6 = 18$
Total number of words $= 1680 + 756 + 18 = 2454$

Stream:JEESubject:MathematicsTopic:Permutation & CombinationSubtopic:nPr & nCr

⏱ 2mℹ️ Source: PYQ_2020

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