permutations

in how many ways can the word "CARTE BLANCHE" can be permutated with no A or E comming adjacently

#Permutation #HelpMe!

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

For just the word CARTE, there are 5! ways to arrange the 5 letters, or 120 ways. If we want the letters A and E to always be adjacentto each other than treat them both as 1 letter. i.e ( AE and EA is the same). Now, there are only 4 letters ( C,AE/EA,R,T) that can be arrange or 4! ways. 4! is 24. Since we want to know how many ways CARTE can be pemutated with no A or E coming adjacently, simply take 120-24= 96 ways.

Venture HI - 7 years, 4 months ago

if u write AECRTE and in similar way u write EACRTE. according to u when we are taking AE as single word EA has no importance . which is 100% wrong because when we arranging these 5 letters in 5! ways both of the above will come ..

so according to me it should be 4!X2! =48.. so total no of permutations is 120-48=72.

Nishant Sharma - 7 years, 4 months ago

You can do the same with BLANCHE using the same thinking...

Venture HI - 7 years, 4 months ago

arrangement of word banana without repition

Kiran Shaji - 5 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$